\displaystyle \textbf{Exercise - 23(A)}

\displaystyle \textbf{Question 1: }\text{Find the value of:}
\displaystyle \text{(i) }\sin30^\circ\cos30^\circ
\displaystyle \text{(ii) }\tan30^\circ\tan60^\circ
\displaystyle \text{(iii) }\cos^2 60^\circ+\sin^2 30^\circ
\displaystyle \text{(iv) }\mathrm{cosec}^2 60^\circ-\tan^2 30^\circ
\displaystyle \text{(v) }\sin^2 30^\circ+\cos^2 30^\circ+\cot^2 45^\circ
\displaystyle \text{(vi) }\cos^2 60^\circ+\sec^2 30^\circ+\tan^2 45^\circ
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\sin30^\circ\cos30^\circ
\displaystyle =\frac{1}{2}\times\frac{\sqrt3}{2}=\frac{\sqrt3}{4}
\displaystyle \therefore \sin30^\circ\cos30^\circ=\frac{\sqrt3}{4}
\displaystyle \text{(ii) }\tan30^\circ\tan60^\circ
\displaystyle =\frac{1}{\sqrt3}\times\sqrt3=1
\displaystyle \therefore \tan30^\circ\tan60^\circ=1
\displaystyle \text{(iii) }\cos^2 60^\circ+\sin^2 30^\circ
\displaystyle =\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2
\displaystyle =\frac14+\frac14=\frac12
\displaystyle \therefore \cos^2 60^\circ+\sin^2 30^\circ=\frac12
\displaystyle \text{(iv) }\mathrm{cosec}^2 60^\circ-\tan^2 30^\circ
\displaystyle =\left(\frac{2}{\sqrt3}\right)^2-\left(\frac{1}{\sqrt3}\right)^2
\displaystyle =\frac43-\frac13=1
\displaystyle \therefore \mathrm{cosec}^2 60^\circ-\tan^2 30^\circ=1
\displaystyle \text{(v) }\sin^2 30^\circ+\cos^2 30^\circ+\cot^2 45^\circ
\displaystyle =\left(\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2+(1)^2
\displaystyle =\frac14+\frac34+1=2
\displaystyle \therefore \sin^2 30^\circ+\cos^2 30^\circ+\cot^2 45^\circ=2
\displaystyle \text{(vi) }\cos^2 60^\circ+\sec^2 30^\circ+\tan^2 45^\circ
\displaystyle =\left(\frac12\right)^2+\left(\frac{2}{\sqrt3}\right)^2+(1)^2
\displaystyle =\frac14+\frac43+1
\displaystyle =\frac{3+16+12}{12}=\frac{31}{12}
\displaystyle \therefore \cos^2 60^\circ+\sec^2 30^\circ+\tan^2 45^\circ=\frac{31}{12}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the value of:}
\displaystyle \text{(i) }\tan^2 30^\circ+\tan^2 45^\circ+\tan^2 60^\circ
\displaystyle \text{(ii) }\frac{\tan45^\circ}{\mathrm{cosec}\,30^\circ}+\frac{\sec60^\circ}{\cot45^\circ}-\frac{5\sin90^\circ}{2\cos0^\circ}
\displaystyle \text{(iii) }3\sin^2 30^\circ+2\tan^2 60^\circ-5\cos^2 45^\circ
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\tan^2 30^\circ+\tan^2 45^\circ+\tan^2 60^\circ
\displaystyle =\left(\frac{1}{\sqrt3}\right)^2+(1)^2+(\sqrt3)^2
\displaystyle =\frac13+1+3=\frac{13}{3}
\displaystyle \therefore \tan^2 30^\circ+\tan^2 45^\circ+\tan^2 60^\circ=\frac{13}{3}
\displaystyle \text{(ii) }\frac{\tan45^\circ}{\mathrm{cosec}\,30^\circ}+\frac{\sec60^\circ}{\cot45^\circ}-\frac{5\sin90^\circ}{2\cos0^\circ}
\displaystyle =\frac{1}{2}+\frac{2}{1}-\frac{5\times1}{2\times1}
\displaystyle =\frac12+2-\frac52=0
\displaystyle \therefore \frac{\tan45^\circ}{\mathrm{cosec}\,30^\circ}+\frac{\sec60^\circ}{\cot45^\circ}-\frac{5\sin90^\circ}{2\cos0^\circ}=0
\displaystyle \text{(iii) }3\sin^2 30^\circ+2\tan^2 60^\circ-5\cos^2 45^\circ
\displaystyle =3\left(\frac12\right)^2+2(\sqrt3)^2-5\left(\frac{1}{\sqrt2}\right)^2
\displaystyle =\frac34+6-\frac52
\displaystyle =\frac{3+24-10}{4}=\frac{17}{4}
\displaystyle \therefore 3\sin^2 30^\circ+2\tan^2 60^\circ-5\cos^2 45^\circ=\frac{17}{4}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Prove that:}
\displaystyle \text{(i) }\sin60^\circ\cos30^\circ+\cos60^\circ\sin30^\circ=1
\displaystyle \text{(ii) }\cos30^\circ\cos60^\circ-\sin30^\circ\sin60^\circ=0
\displaystyle \text{(iii) }\mathrm{cosec}^2 45^\circ-\cot^2 45^\circ=1
\displaystyle \text{(iv) }\cos^2 30^\circ-\sin^2 30^\circ=\cos60^\circ
\displaystyle \text{(v) }\left(\frac{\tan60^\circ+1}{\tan60^\circ-1}\right)^2=\frac{1+\cos30^\circ}{1-\cos30^\circ}
\displaystyle \text{(vi) }3\mathrm{cosec}^2 60^\circ-2\cot^2 30^\circ+\sec^2 45^\circ=0
\displaystyle \text{Answer:}
\displaystyle \text{(i) L.H.S.}=\sin60^\circ\cos30^\circ+\cos60^\circ\sin30^\circ
\displaystyle =\frac{\sqrt3}{2}\times\frac{\sqrt3}{2}+\frac12\times\frac12
\displaystyle =\frac34+\frac14=1=\text{R.H.S.}
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}

\displaystyle \text{(ii) L.H.S.}=\cos30^\circ\cos60^\circ-\sin30^\circ\sin60^\circ
\displaystyle =\frac{\sqrt3}{2}\times\frac12-\frac12\times\frac{\sqrt3}{2}
\displaystyle =\frac{\sqrt3}{4}-\frac{\sqrt3}{4}=0=\text{R.H.S.}
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}

\displaystyle \text{(iii) L.H.S.}=\mathrm{cosec}^2 45^\circ-\cot^2 45^\circ
\displaystyle =(\sqrt2)^2-(1)^2
\displaystyle =2-1=1=\text{R.H.S.}
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}

\displaystyle \text{(iv) L.H.S.}=\cos^2 30^\circ-\sin^2 30^\circ
\displaystyle =\left(\frac{\sqrt3}{2}\right)^2-\left(\frac12\right)^2
\displaystyle =\frac34-\frac14=\frac12
\displaystyle \text{R.H.S.}=\cos60^\circ=\frac12
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}

\displaystyle \text{(v) L.H.S.}=\left(\frac{\tan60^\circ+1}{\tan60^\circ-1}\right)^2
\displaystyle =\left(\frac{\sqrt3+1}{\sqrt3-1}\right)^2
\displaystyle =\left(\frac{(\sqrt3+1)^2}{3-1}\right)^2
\displaystyle =\left(\frac{4+2\sqrt3}{2}\right)^2=(2+\sqrt3)^2
\displaystyle =7+4\sqrt3
\displaystyle \text{R.H.S.}=\frac{1+\cos30^\circ}{1-\cos30^\circ}
\displaystyle =\frac{1+\frac{\sqrt3}{2}}{1-\frac{\sqrt3}{2}}=\frac{2+\sqrt3}{2-\sqrt3}
\displaystyle =\frac{(2+\sqrt3)^2}{4-3}=(2+\sqrt3)^2
\displaystyle =7+4\sqrt3
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}

\displaystyle \text{(vi) L.H.S.}=3\mathrm{cosec}^2 60^\circ-2\cot^2 30^\circ+\sec^2 45^\circ
\displaystyle =3\left(\frac{2}{\sqrt3}\right)^2-2(\sqrt3)^2+(\sqrt2)^2
\displaystyle =3\times\frac43-2\times3+2
\displaystyle =4-6+2=0=\text{R.H.S.}
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Prove that:}
\displaystyle \text{(i) }\sin(2\times30^\circ)=\frac{2\tan30^\circ}{1+\tan^2 30^\circ}
\displaystyle \text{(ii) }\cos(2\times30^\circ)=\frac{1-\tan^2 30^\circ}{1+\tan^2 30^\circ}
\displaystyle \text{(iii) }\tan(2\times30^\circ)=\frac{2\tan30^\circ}{1-\tan^2 30^\circ}
\displaystyle \text{Answer:}
\displaystyle \text{(i) L.H.S.}=\sin(2\times30^\circ)=\sin60^\circ=\frac{\sqrt3}{2}
\displaystyle \text{R.H.S.}=\frac{2\tan30^\circ}{1+\tan^2 30^\circ}
\displaystyle =\frac{2\times\frac{1}{\sqrt3}}{1+\left(\frac{1}{\sqrt3}\right)^2}
\displaystyle =\frac{\frac{2}{\sqrt3}}{1+\frac13}=\frac{\frac{2}{\sqrt3}}{\frac43}
\displaystyle =\frac{2}{\sqrt3}\times\frac34=\frac{\sqrt3}{2}
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}

\displaystyle \text{(ii) L.H.S.}=\cos(2\times30^\circ)=\cos60^\circ=\frac12
\displaystyle \text{R.H.S.}=\frac{1-\tan^2 30^\circ}{1+\tan^2 30^\circ}
\displaystyle =\frac{1-\left(\frac{1}{\sqrt3}\right)^2}{1+\left(\frac{1}{\sqrt3}\right)^2}
\displaystyle =\frac{1-\frac13}{1+\frac13}=\frac{\frac23}{\frac43}=\frac12
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}

\displaystyle \text{(iii) L.H.S.}=\tan(2\times30^\circ)=\tan60^\circ=\sqrt3
\displaystyle \text{R.H.S.}=\frac{2\tan30^\circ}{1-\tan^2 30^\circ}
\displaystyle =\frac{2\times\frac{1}{\sqrt3}}{1-\left(\frac{1}{\sqrt3}\right)^2}
\displaystyle =\frac{\frac{2}{\sqrt3}}{1-\frac13}=\frac{\frac{2}{\sqrt3}}{\frac23}
\displaystyle =\frac{2}{\sqrt3}\times\frac32=\sqrt3
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{ABC is an isosceles right-angled triangle. Assuming }AB=BC=x,
\displaystyle \text{find the value of each of the following trigonometric ratios:}
\displaystyle \text{(i) }\sin45^\circ\qquad\text{(ii) }\cos45^\circ\qquad\text{(iii) }\tan45^\circ
\displaystyle \text{Answer:}
\displaystyle \text{Given, }AB=BC=x\text{ and }\angle B=90^\circ.
\displaystyle \therefore \angle A=\angle C=45^\circ
\displaystyle \text{By Pythagoras theorem,}
\displaystyle AC^2=AB^2+BC^2=x^2+x^2=2x^2
\displaystyle \therefore AC=x\sqrt2
\displaystyle \text{(i) }\sin45^\circ=\frac{BC}{AC}=\frac{x}{x\sqrt2}=\frac{1}{\sqrt2}
\displaystyle \text{(ii) }\cos45^\circ=\frac{AB}{AC}=\frac{x}{x\sqrt2}=\frac{1}{\sqrt2}
\displaystyle \text{(iii) }\tan45^\circ=\frac{BC}{AB}=\frac{x}{x}=1
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Prove that:}
\displaystyle \text{(i) }\sin60^\circ=2\sin30^\circ\cos30^\circ
\displaystyle \text{(ii) }4(\sin^4 30^\circ+\cos^4 60^\circ)-3(\cos^2 45^\circ-\sin^2 90^\circ)=2
\displaystyle \text{Answer:}
\displaystyle \text{(i) L.H.S.}=\sin60^\circ=\frac{\sqrt3}{2}
\displaystyle \text{R.H.S.}=2\sin30^\circ\cos30^\circ
\displaystyle =2\times\frac12\times\frac{\sqrt3}{2}=\frac{\sqrt3}{2}
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}

\displaystyle \text{(ii) L.H.S.}=4(\sin^4 30^\circ+\cos^4 60^\circ)-3(\cos^2 45^\circ-\sin^2 90^\circ)
\displaystyle =4\left[\left(\frac12\right)^4+\left(\frac12\right)^4\right]-3\left[\left(\frac{1}{\sqrt2}\right)^2-(1)^2\right]
\displaystyle =4\left(\frac{1}{16}+\frac{1}{16}\right)-3\left(\frac12-1\right)
\displaystyle =4\times\frac18-3\left(-\frac12\right)
\displaystyle =\frac12+\frac32=2=\text{R.H.S.}
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{(i) If }\sin x=\cos x\text{ and }x\text{ is acute, state the value of }x.
\displaystyle \text{(ii) If }\sec A=\mathrm{cosec}\,A\text{ and }0^\circ\leq A\leq90^\circ,\text{ state the value of }A.
\displaystyle \text{(iii) If }\tan\theta=\cot\theta\text{ and }0^\circ\leq\theta\leq90^\circ,\text{ state the value of }\theta.
\displaystyle \text{(iv) If }\sin x=\cos y,\text{ write the relation between }x\text{ and }y,\text{ if both angles are acute.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given, }\sin x=\cos x
\displaystyle \sin45^\circ=\cos45^\circ=\frac{1}{\sqrt2}
\displaystyle \therefore x=45^\circ

\displaystyle \text{(ii) Given, }\sec A=\mathrm{cosec}\,A
\displaystyle \sec45^\circ=\mathrm{cosec}\,45^\circ=\sqrt2
\displaystyle \therefore A=45^\circ

\displaystyle \text{(iii) Given, }\tan\theta=\cot\theta
\displaystyle \tan45^\circ=\cot45^\circ=1
\displaystyle \therefore \theta=45^\circ

\displaystyle \text{(iv) Given, }\sin x=\cos y
\displaystyle \text{For acute complementary angles, }\sin x=\cos(90^\circ-x)
\displaystyle \therefore y=90^\circ-x
\displaystyle \therefore x+y=90^\circ
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{(i) If }\sin x=\cos y,\text{ then }x+y=45^\circ;\text{ write true or false.}
\displaystyle \text{(ii) }\sec\theta\cot\theta=\mathrm{cosec}\,\theta;\text{ write true or false.}
\displaystyle \text{(iii) For any angle }\theta,\text{ state the value of }\sin^2\theta+\cos^2\theta.
\displaystyle \text{Answer:}
\displaystyle \text{(i) False.}
\displaystyle \text{For acute angles, }\sin x=\cos y\Rightarrow x+y=90^\circ.

\displaystyle \text{(ii) True.}
\displaystyle \sec\theta\cot\theta=\frac{1}{\cos\theta}\times\frac{\cos\theta}{\sin\theta}
\displaystyle =\frac{1}{\sin\theta}=\mathrm{cosec}\,\theta

\displaystyle \text{(iii) }\sin^2\theta+\cos^2\theta=1
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{State for any acute angle }\theta\text{ whether:}
\displaystyle \text{(i) }\sin\theta\text{ increases or decreases as }\theta\text{ increases.}
\displaystyle \text{(ii) }\cos\theta\text{ increases or decreases as }\theta\text{ increases.}
\displaystyle \text{(iii) }\tan\theta\text{ increases or decreases as }\theta\text{ decreases.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\sin\theta\text{ increases as }\theta\text{ increases.}
\displaystyle \text{(ii) }\cos\theta\text{ decreases as }\theta\text{ increases.}
\displaystyle \text{(iii) }\tan\theta\text{ decreases as }\theta\text{ decreases.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }\sqrt3=1.732,\text{ find (correct to two decimal places) the value of}
\displaystyle \text{each of the following:}
\displaystyle \text{(i) }\sin60^\circ\qquad\text{(ii) }\frac{2}{\tan30^\circ}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\sin60^\circ=\frac{\sqrt3}{2}
\displaystyle =\frac{1.732}{2}=0.866
\displaystyle \therefore \sin60^\circ=0.87\text{ (correct to two decimal places).}

\displaystyle \text{(ii) }\frac{2}{\tan30^\circ}=\frac{2}{\frac{1}{\sqrt3}}
\displaystyle =2\sqrt3=2\times1.732=3.464
\displaystyle \therefore \frac{2}{\tan30^\circ}=3.46\text{ (correct to two decimal places).}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Evaluate:}
\displaystyle \text{(i) }\frac{\cos3A-2\cos4A}{\sin3A+2\sin4A},\text{ when }A=15^\circ.
\displaystyle \text{(ii) }\frac{3\sin3B+2\cos(2B+5^\circ)}{2\cos3B-\sin(2B-10^\circ)},\text{ when }B=20^\circ.
\displaystyle \text{Answer:}
\displaystyle \text{(i) When }A=15^\circ,\quad 3A=45^\circ\text{ and }4A=60^\circ.
\displaystyle \frac{\cos3A-2\cos4A}{\sin3A+2\sin4A}
\displaystyle =\frac{\cos45^\circ-2\cos60^\circ}{\sin45^\circ+2\sin60^\circ}
\displaystyle =\frac{\frac{1}{\sqrt2}-2\times\frac12}{\frac{1}{\sqrt2}+2\times\frac{\sqrt3}{2}}
\displaystyle =\frac{\frac{1}{\sqrt2}-1}{\frac{1}{\sqrt2}+\sqrt3}

\displaystyle \text{(ii) When }B=20^\circ,\quad 3B=60^\circ,\quad 2B+5^\circ=45^\circ,\quad 2B-10^\circ=30^\circ.
\displaystyle \frac{3\sin3B+2\cos(2B+5^\circ)}{2\cos3B-\sin(2B-10^\circ)}
\displaystyle =\frac{3\sin60^\circ+2\cos45^\circ}{2\cos60^\circ-\sin30^\circ}
\displaystyle =\frac{3\times\frac{\sqrt3}{2}+2\times\frac{1}{\sqrt2}}{2\times\frac12-\frac12}
\displaystyle =\frac{\frac{3\sqrt3}{2}+\sqrt2}{\frac12}
\displaystyle =3\sqrt3+2\sqrt2
\displaystyle \\

\displaystyle \textbf{Exercise - 23(B)}

\displaystyle \textbf{Question 1: }\text{Given }A=60^\circ\text{ and }B=30^\circ,\text{ prove that:}
\displaystyle \text{(i) }\sin(A+B)=\sin A\cos B+\cos A\sin B
\displaystyle \text{(ii) }\cos(A+B)=\cos A\cos B-\sin A\sin B
\displaystyle \text{(iii) }\cos(A-B)=\cos A\cos B+\sin A\sin B
\displaystyle \text{(iv) }\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=60^\circ\text{ and }B=30^\circ.
\displaystyle \text{(i) L.H.S.}=\sin(A+B)=\sin(60^\circ+30^\circ)
\displaystyle =\sin90^\circ=1
\displaystyle \text{R.H.S.}=\sin A\cos B+\cos A\sin B
\displaystyle =\sin60^\circ\cos30^\circ+\cos60^\circ\sin30^\circ
\displaystyle =\frac{\sqrt3}{2}\times\frac{\sqrt3}{2}+\frac12\times\frac12
\displaystyle =\frac34+\frac14=1
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}

\displaystyle \text{(ii) L.H.S.}=\cos(A+B)=\cos(60^\circ+30^\circ)
\displaystyle =\cos90^\circ=0
\displaystyle \text{R.H.S.}=\cos A\cos B-\sin A\sin B
\displaystyle =\cos60^\circ\cos30^\circ-\sin60^\circ\sin30^\circ
\displaystyle =\frac12\times\frac{\sqrt3}{2}-\frac{\sqrt3}{2}\times\frac12
\displaystyle =\frac{\sqrt3}{4}-\frac{\sqrt3}{4}=0
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}

\displaystyle \text{(iii) L.H.S.}=\cos(A-B)=\cos(60^\circ-30^\circ)
\displaystyle =\cos30^\circ=\frac{\sqrt3}{2}
\displaystyle \text{R.H.S.}=\cos A\cos B+\sin A\sin B
\displaystyle =\cos60^\circ\cos30^\circ+\sin60^\circ\sin30^\circ
\displaystyle =\frac12\times\frac{\sqrt3}{2}+\frac{\sqrt3}{2}\times\frac12
\displaystyle =\frac{\sqrt3}{4}+\frac{\sqrt3}{4}=\frac{\sqrt3}{2}
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}

\displaystyle \text{(iv) L.H.S.}=\tan(A-B)=\tan(60^\circ-30^\circ)
\displaystyle =\tan30^\circ=\frac{1}{\sqrt3}
\displaystyle \text{R.H.S.}=\frac{\tan A-\tan B}{1+\tan A\tan B}
\displaystyle =\frac{\sqrt3-\frac{1}{\sqrt3}}{1+\sqrt3\times\frac{1}{\sqrt3}}
\displaystyle =\frac{\frac{2}{\sqrt3}}{2}=\frac{1}{\sqrt3}
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }A=30^\circ,\text{ then prove that:}
\displaystyle \text{(i) }\sin2A=2\sin A\cos A=\frac{2\tan A}{1+\tan^2 A}
\displaystyle \text{(ii) }\cos2A=\cos^2 A-\sin^2 A=\frac{1-\tan^2 A}{1+\tan^2 A}
\displaystyle \text{(iii) }2\cos^2 A-1=1-2\sin^2 A
\displaystyle \text{(iv) }\sin3A=3\sin A-4\sin^3 A
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=30^\circ.
\displaystyle \text{(i) L.H.S.}=\sin2A=\sin60^\circ=\frac{\sqrt3}{2}
\displaystyle 2\sin A\cos A=2\sin30^\circ\cos30^\circ
\displaystyle =2\times\frac12\times\frac{\sqrt3}{2}=\frac{\sqrt3}{2}
\displaystyle \frac{2\tan A}{1+\tan^2 A}=\frac{2\tan30^\circ}{1+\tan^2 30^\circ}
\displaystyle =\frac{2\times\frac1{\sqrt3}}{1+\left(\frac1{\sqrt3}\right)^2}
\displaystyle =\frac{\frac2{\sqrt3}}{\frac43}=\frac{\sqrt3}{2}
\displaystyle \therefore \sin2A=2\sin A\cos A=\frac{2\tan A}{1+\tan^2 A}

\displaystyle \text{(ii) L.H.S.}=\cos2A=\cos60^\circ=\frac12
\displaystyle \cos^2 A-\sin^2 A=\cos^2 30^\circ-\sin^2 30^\circ
\displaystyle =\left(\frac{\sqrt3}{2}\right)^2-\left(\frac12\right)^2
\displaystyle =\frac34-\frac14=\frac12
\displaystyle \frac{1-\tan^2 A}{1+\tan^2 A}=\frac{1-\tan^2 30^\circ}{1+\tan^2 30^\circ}
\displaystyle =\frac{1-\frac13}{1+\frac13}=\frac{\frac23}{\frac43}=\frac12
\displaystyle \therefore \cos2A=\cos^2 A-\sin^2 A=\frac{1-\tan^2 A}{1+\tan^2 A}

\displaystyle \text{(iii) L.H.S.}=2\cos^2 A-1
\displaystyle =2\left(\frac{\sqrt3}{2}\right)^2-1
\displaystyle =2\times\frac34-1=\frac12
\displaystyle \text{R.H.S.}=1-2\sin^2 A
\displaystyle =1-2\left(\frac12\right)^2=1-\frac12=\frac12
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}

\displaystyle \text{(iv) L.H.S.}=\sin3A=\sin90^\circ=1
\displaystyle \text{R.H.S.}=3\sin A-4\sin^3 A
\displaystyle =3\left(\frac12\right)-4\left(\frac12\right)^3
\displaystyle =\frac32-\frac12=1
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }A=B=45^\circ,\text{ show that:}
\displaystyle \text{(i) }\sin(A-B)=\sin A\cos B-\cos A\sin B
\displaystyle \text{(ii) }\cos(A+B)=\cos A\cos B-\sin A\sin B
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=B=45^\circ.
\displaystyle \text{(i) L.H.S.}=\sin(A-B)=\sin(45^\circ-45^\circ)
\displaystyle =\sin0^\circ=0
\displaystyle \text{R.H.S.}=\sin A\cos B-\cos A\sin B
\displaystyle =\sin45^\circ\cos45^\circ-\cos45^\circ\sin45^\circ
\displaystyle =\frac{1}{\sqrt2}\times\frac{1}{\sqrt2}-\frac{1}{\sqrt2}\times\frac{1}{\sqrt2}
\displaystyle =\frac12-\frac12=0
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}

\displaystyle \text{(ii) L.H.S.}=\cos(A+B)=\cos(45^\circ+45^\circ)
\displaystyle =\cos90^\circ=0
\displaystyle \text{R.H.S.}=\cos A\cos B-\sin A\sin B
\displaystyle =\cos45^\circ\cos45^\circ-\sin45^\circ\sin45^\circ
\displaystyle =\frac{1}{\sqrt2}\times\frac{1}{\sqrt2}-\frac{1}{\sqrt2}\times\frac{1}{\sqrt2}
\displaystyle =\frac12-\frac12=0
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }A=30^\circ,\text{ show that:}
\displaystyle \text{(i) }\sin3A=4\sin A\sin(60^\circ-A)\sin(60^\circ+A)
\displaystyle \text{(ii) }(\sin A-\cos A)^2=1-\sin2A
\displaystyle \text{(iii) }\cos2A=\cos^4 A-\sin^4 A
\displaystyle \text{(iv) }\frac{1-\cos2A}{\sin2A}=\tan A
\displaystyle \text{(v) }\frac{1+\sin2A+\cos2A}{\sin A+\cos A}=2\cos A
\displaystyle \text{(vi) }4\cos A\cos(60^\circ-A)\cos(60^\circ+A)=\cos3A
\displaystyle \text{(vii) }\frac{\cos^3 A-\cos3A}{\cos A}+\frac{\sin^3 A+\sin3A}{\sin A}=3
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=30^\circ.
\displaystyle \text{(i) L.H.S.}=\sin3A=\sin90^\circ=1
\displaystyle \text{R.H.S.}=4\sin A\sin(60^\circ-A)\sin(60^\circ+A)
\displaystyle =4\sin30^\circ\sin30^\circ\sin90^\circ
\displaystyle =4\times\frac12\times\frac12\times1=1
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}

\displaystyle \text{(ii) L.H.S.}=(\sin A-\cos A)^2
\displaystyle =\left(\frac12-\frac{\sqrt3}{2}\right)^2
\displaystyle =\frac{(1-\sqrt3)^2}{4}=1-\frac{\sqrt3}{2}
\displaystyle \text{R.H.S.}=1-\sin2A=1-\sin60^\circ
\displaystyle =1-\frac{\sqrt3}{2}
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}

\displaystyle \text{(iii) L.H.S.}=\cos2A=\cos60^\circ=\frac12
\displaystyle \text{R.H.S.}=\cos^4 A-\sin^4 A
\displaystyle =\left(\frac{\sqrt3}{2}\right)^4-\left(\frac12\right)^4
\displaystyle =\frac9{16}-\frac1{16}=\frac12
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}

\displaystyle \text{(iv) L.H.S.}=\frac{1-\cos2A}{\sin2A}
\displaystyle =\frac{1-\cos60^\circ}{\sin60^\circ}
\displaystyle =\frac{1-\frac12}{\frac{\sqrt3}{2}}=\frac1{\sqrt3}
\displaystyle \text{R.H.S.}=\tan A=\tan30^\circ=\frac1{\sqrt3}
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}

\displaystyle \text{(v) L.H.S.}=\frac{1+\sin2A+\cos2A}{\sin A+\cos A}
\displaystyle =\frac{1+\sin60^\circ+\cos60^\circ}{\sin30^\circ+\cos30^\circ}
\displaystyle =\frac{1+\frac{\sqrt3}{2}+\frac12}{\frac12+\frac{\sqrt3}{2}}
\displaystyle =\frac{3+\sqrt3}{1+\sqrt3}
\displaystyle =\frac{\sqrt3(1+\sqrt3)}{1+\sqrt3}=\sqrt3
\displaystyle \text{R.H.S.}=2\cos A=2\cos30^\circ
\displaystyle =2\times\frac{\sqrt3}{2}=\sqrt3
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}

\displaystyle \text{(vi) L.H.S.}=4\cos A\cos(60^\circ-A)\cos(60^\circ+A)
\displaystyle =4\cos30^\circ\cos30^\circ\cos90^\circ
\displaystyle =4\times\frac{\sqrt3}{2}\times\frac{\sqrt3}{2}\times0=0
\displaystyle \text{R.H.S.}=\cos3A=\cos90^\circ=0
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}

\displaystyle \text{(vii) L.H.S.}=\frac{\cos^3 A-\cos3A}{\cos A}+\frac{\sin^3 A+\sin3A}{\sin A}
\displaystyle =\frac{\left(\frac{\sqrt3}{2}\right)^3-\cos90^\circ}{\frac{\sqrt3}{2}}+\frac{\left(\frac12\right)^3+\sin90^\circ}{\frac12}
\displaystyle =\frac{\frac{3\sqrt3}{8}}{\frac{\sqrt3}{2}}+\frac{\frac18+1}{\frac12}
\displaystyle =\frac34+\frac94=3=\text{R.H.S.}
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Exercise - 23(C)}

\displaystyle \textbf{Question 1: }\text{Solve the following equations for }A,\text{ if:}
\displaystyle \text{(i) }2\sin A=1\qquad\text{(ii) }2\cos2A=1
\displaystyle \text{(iii) }\sin3A=\frac{\sqrt3}{2}\qquad\text{(iv) }\sec2A=2
\displaystyle \text{(v) }\sqrt3\tan A=1\qquad\text{(vi) }\tan3A=1
\displaystyle \text{(vii) }2\sin3A=1\qquad\text{(viii) }\sqrt3\cot2A=1
\displaystyle \text{Answer:}
\displaystyle \text{(i) }2\sin A=1
\displaystyle \Rightarrow \sin A=\frac12=\sin30^\circ
\displaystyle \therefore A=30^\circ

\displaystyle \text{(ii) }2\cos2A=1
\displaystyle \Rightarrow \cos2A=\frac12=\cos60^\circ
\displaystyle \Rightarrow 2A=60^\circ
\displaystyle \therefore A=30^\circ

\displaystyle \text{(iii) }\sin3A=\frac{\sqrt3}{2}=\sin60^\circ
\displaystyle \Rightarrow 3A=60^\circ
\displaystyle \therefore A=20^\circ

\displaystyle \text{(iv) }\sec2A=2
\displaystyle \Rightarrow \sec2A=\sec60^\circ
\displaystyle \Rightarrow 2A=60^\circ
\displaystyle \therefore A=30^\circ

\displaystyle \text{(v) }\sqrt3\tan A=1
\displaystyle \Rightarrow \tan A=\frac1{\sqrt3}=\tan30^\circ
\displaystyle \therefore A=30^\circ

\displaystyle \text{(vi) }\tan3A=1=\tan45^\circ
\displaystyle \Rightarrow 3A=45^\circ
\displaystyle \therefore A=15^\circ

\displaystyle \text{(vii) }2\sin3A=1
\displaystyle \Rightarrow \sin3A=\frac12=\sin30^\circ
\displaystyle \Rightarrow 3A=30^\circ
\displaystyle \therefore A=10^\circ

\displaystyle \text{(viii) }\sqrt3\cot2A=1
\displaystyle \Rightarrow \cot2A=\frac1{\sqrt3}=\cot60^\circ
\displaystyle \Rightarrow 2A=60^\circ
\displaystyle \therefore A=30^\circ
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Calculate the value of }A,\text{ if:}
\displaystyle \text{(i) }(\sin A-1)(2\cos A-1)=0
\displaystyle \text{(ii) }(\tan A-1)(\mathrm{cosec}\,3A-1)=0
\displaystyle \text{(iii) }(\sec2A-1)(\mathrm{cosec}\,3A-1)=0
\displaystyle \text{(iv) }\cos3A(2\sin2A-1)=0
\displaystyle \text{(v) }(\mathrm{cosec}\,2A-2)(\cot3A-1)=0
\displaystyle \text{Answer:}
\displaystyle \text{(i) }(\sin A-1)(2\cos A-1)=0
\displaystyle \Rightarrow \sin A-1=0\quad\text{or}\quad2\cos A-1=0
\displaystyle \Rightarrow \sin A=1\quad\text{or}\quad\cos A=\frac12
\displaystyle \Rightarrow \sin A=\sin90^\circ\quad\text{or}\quad\cos A=\cos60^\circ
\displaystyle \therefore A=90^\circ\text{ or }60^\circ

\displaystyle \text{(ii) }(\tan A-1)(\mathrm{cosec}\,3A-1)=0
\displaystyle \Rightarrow \tan A=1\quad\text{or}\quad\mathrm{cosec}\,3A=1
\displaystyle \Rightarrow \tan A=\tan45^\circ\quad\text{or}\quad\sin3A=1
\displaystyle \Rightarrow A=45^\circ\quad\text{or}\quad3A=90^\circ
\displaystyle \therefore A=45^\circ\text{ or }30^\circ

\displaystyle \text{(iii) }(\sec2A-1)(\mathrm{cosec}\,3A-1)=0
\displaystyle \Rightarrow \sec2A=1\quad\text{or}\quad\mathrm{cosec}\,3A=1
\displaystyle \Rightarrow \cos2A=1\quad\text{or}\quad\sin3A=1
\displaystyle \Rightarrow 2A=0^\circ\quad\text{or}\quad3A=90^\circ
\displaystyle \therefore A=0^\circ\text{ or }30^\circ

\displaystyle \text{(iv) }\cos3A(2\sin2A-1)=0
\displaystyle \Rightarrow \cos3A=0\quad\text{or}\quad2\sin2A-1=0
\displaystyle \Rightarrow \cos3A=\cos90^\circ\quad\text{or}\quad\sin2A=\frac12
\displaystyle \Rightarrow 3A=90^\circ\quad\text{or}\quad2A=30^\circ
\displaystyle \therefore A=30^\circ\text{ or }15^\circ

\displaystyle \text{(v) }(\mathrm{cosec}\,2A-2)(\cot3A-1)=0
\displaystyle \Rightarrow \mathrm{cosec}\,2A=2\quad\text{or}\quad\cot3A=1
\displaystyle \Rightarrow \sin2A=\frac12\quad\text{or}\quad\cot3A=\cot45^\circ
\displaystyle \Rightarrow 2A=30^\circ\quad\text{or}\quad3A=45^\circ
\displaystyle \therefore A=15^\circ
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }2\sin x^\circ-1=0\text{ and }x^\circ\text{ is an acute angle, find:}
\displaystyle \text{(i) }\sin x^\circ\qquad\text{(ii) }x^\circ\qquad\text{(iii) }\cos x^\circ\text{ and }\tan x^\circ.
\displaystyle \text{Answer:}
\displaystyle 2\sin x^\circ-1=0
\displaystyle \Rightarrow 2\sin x^\circ=1
\displaystyle \Rightarrow \sin x^\circ=\frac12
\displaystyle \text{(i) }\therefore \sin x^\circ=\frac12

\displaystyle \text{(ii) Since }\sin30^\circ=\frac12,
\displaystyle \therefore x^\circ=30^\circ

\displaystyle \text{(iii) }\cos x^\circ=\cos30^\circ=\frac{\sqrt3}{2}
\displaystyle \tan x^\circ=\tan30^\circ=\frac1{\sqrt3}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }4\cos^2 x^\circ-1=0\text{ and }0\leq x^\circ\leq90^\circ,\text{ find:}
\displaystyle \text{(i) }x^\circ\qquad\text{(ii) }\sin^2 x^\circ+\cos^2 x^\circ
\displaystyle \text{(iii) }\frac{1}{\cos^2 x^\circ}-\tan^2 x^\circ
\displaystyle \text{Answer:}
\displaystyle 4\cos^2 x^\circ-1=0
\displaystyle \Rightarrow 4\cos^2 x^\circ=1
\displaystyle \Rightarrow \cos^2 x^\circ=\frac14
\displaystyle \Rightarrow \cos x^\circ=\frac12
\displaystyle \text{(Since }0^\circ\leq x^\circ\leq90^\circ,\ \cos x^\circ\geq0\text{.)}
\displaystyle \text{(i) Since }\cos60^\circ=\frac12,
\displaystyle \therefore x^\circ=60^\circ

\displaystyle \text{(ii) }\sin^2 x^\circ+\cos^2 x^\circ
\displaystyle =\sin^2 60^\circ+\cos^2 60^\circ
\displaystyle =\left(\frac{\sqrt3}{2}\right)^2+\left(\frac12\right)^2
\displaystyle =\frac34+\frac14=1

\displaystyle \text{(iii) }\frac{1}{\cos^2 x^\circ}-\tan^2 x^\circ
\displaystyle =\frac{1}{\cos^2 60^\circ}-\tan^2 60^\circ
\displaystyle =\frac{1}{\left(\frac12\right)^2}-(\sqrt3)^2
\displaystyle =4-3=1
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }4\sin^2\theta-1=0\text{ and angle }\theta\text{ is less than }90^\circ,\text{ find the}
\displaystyle \text{value of }\theta\text{ and hence the value of }\cos^2\theta+\tan^2\theta.
\displaystyle \text{Answer:}
\displaystyle 4\sin^2\theta-1=0
\displaystyle \Rightarrow 4\sin^2\theta=1
\displaystyle \Rightarrow \sin^2\theta=\frac14
\displaystyle \Rightarrow \sin\theta=\frac12
\displaystyle \text{Since }\sin30^\circ=\frac12,
\displaystyle \therefore \theta=30^\circ
\displaystyle \cos^2\theta+\tan^2\theta
\displaystyle =\cos^2 30^\circ+\tan^2 30^\circ
\displaystyle =\left(\frac{\sqrt3}{2}\right)^2+\left(\frac1{\sqrt3}\right)^2
\displaystyle =\frac34+\frac13=\frac{9+4}{12}=\frac{13}{12}
\displaystyle \therefore \cos^2\theta+\tan^2\theta=\frac{13}{12}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }\sin3A=1\text{ and }0^\circ\leq A\leq90^\circ,\text{ find:}
\displaystyle \text{(i) }\sin A\qquad\text{(ii) }\cos2A\qquad\text{(iii) }\tan^2 A-\frac{1}{\cos^2 A}
\displaystyle \text{Answer:}
\displaystyle \sin3A=1=\sin90^\circ
\displaystyle \Rightarrow 3A=90^\circ
\displaystyle \therefore A=30^\circ
\displaystyle \text{(i) }\sin A=\sin30^\circ=\frac12

\displaystyle \text{(ii) }\cos2A=\cos60^\circ=\frac12

\displaystyle \text{(iii) }\tan^2 A-\frac{1}{\cos^2 A}
\displaystyle =\tan^2 30^\circ-\frac{1}{\cos^2 30^\circ}
\displaystyle =\left(\frac1{\sqrt3}\right)^2-\frac{1}{\left(\frac{\sqrt3}{2}\right)^2}
\displaystyle =\frac13-\frac43=-1
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If }2\cos2A=\sqrt3\text{ and }A\text{ is acute, find:}
\displaystyle \text{(i) }A\qquad\text{(ii) }\sin3A
\displaystyle \text{(iii) }\sin^2(75^\circ-A)+\cos^2(45^\circ+A)
\displaystyle \text{Answer:}
\displaystyle 2\cos2A=\sqrt3
\displaystyle \Rightarrow \cos2A=\frac{\sqrt3}{2}=\cos30^\circ
\displaystyle \Rightarrow 2A=30^\circ
\displaystyle \text{(i) }\therefore A=15^\circ

\displaystyle \text{(ii) }\sin3A=\sin(3\times15^\circ)
\displaystyle =\sin45^\circ=\frac1{\sqrt2}

\displaystyle \text{(iii) }\sin^2(75^\circ-A)+\cos^2(45^\circ+A)
\displaystyle =\sin^2(75^\circ-15^\circ)+\cos^2(45^\circ+15^\circ)
\displaystyle =\sin^2 60^\circ+\cos^2 60^\circ
\displaystyle =\left(\frac{\sqrt3}{2}\right)^2+\left(\frac12\right)^2
\displaystyle =\frac34+\frac14=1
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{(i) If }\sin x+\cos y=1\text{ and }x=30^\circ,\text{ find the value of }y.
\displaystyle \text{(ii) If }3\tan A-5\cos B=\sqrt3\text{ and }B=90^\circ,\text{ find the value of }A.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given, }\sin x+\cos y=1\text{ and }x=30^\circ.
\displaystyle \therefore \sin30^\circ+\cos y=1
\displaystyle \Rightarrow \frac12+\cos y=1
\displaystyle \Rightarrow \cos y=\frac12
\displaystyle \Rightarrow \cos y=\cos60^\circ
\displaystyle \therefore y=60^\circ

\displaystyle \text{(ii) Given, }3\tan A-5\cos B=\sqrt3\text{ and }B=90^\circ.
\displaystyle \therefore 3\tan A-5\cos90^\circ=\sqrt3
\displaystyle \Rightarrow 3\tan A=\sqrt3
\displaystyle \Rightarrow \tan A=\frac{\sqrt3}{3}=\frac1{\sqrt3}
\displaystyle \Rightarrow \tan A=\tan30^\circ
\displaystyle \therefore A=30^\circ
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{From the given figure, find:}
\displaystyle \text{(i) }\cos x^\circ\qquad\text{(ii) }x^\circ
\displaystyle \text{(iii) }\frac{1}{\tan^2 x^\circ}-\frac{1}{\sin^2 x^\circ}
\displaystyle \text{(iv) Use }x^\circ\text{ to find the value of }y.
\displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle ABC,\quad AC=20\text{ cm},\quad BC=10\text{ cm}.
\displaystyle \text{(i) }\cos x^\circ=\frac{\text{base}}{\text{hypotenuse}}=\frac{BC}{AC}
\displaystyle =\frac{10}{20}=\frac12

\displaystyle \text{(ii) Since }\cos x^\circ=\frac12=\cos60^\circ,
\displaystyle \therefore x^\circ=60^\circ

\displaystyle \text{(iii) }\frac{1}{\tan^2 x^\circ}-\frac{1}{\sin^2 x^\circ}
\displaystyle =\frac{1}{\tan^2 60^\circ}-\frac{1}{\sin^2 60^\circ}
\displaystyle =\frac{1}{(\sqrt3)^2}-\frac{1}{\left(\frac{\sqrt3}{2}\right)^2}
\displaystyle =\frac13-\frac43=-1

\displaystyle \text{(iv) }\tan x^\circ=\frac{AB}{BC}=\frac{y}{10}
\displaystyle \therefore \tan60^\circ=\frac{y}{10}
\displaystyle \Rightarrow \sqrt3=\frac{y}{10}
\displaystyle \therefore y=10\sqrt3\text{ cm}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Use the given figure to find:}
\displaystyle \text{(i) }\tan\theta^\circ\qquad\text{(ii) }\theta^\circ\qquad\text{(iii) }\sin^2\theta^\circ-\cos^2\theta^\circ
\displaystyle \text{(iv) Use }\sin\theta^\circ\text{ to find the value of }x.
\displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle ABC,\quad AB=BC=5.
\displaystyle \text{(i) }\tan\theta^\circ=\frac{AB}{BC}=\frac55=1

\displaystyle \text{(ii) Since }\tan\theta^\circ=1=\tan45^\circ,
\displaystyle \therefore \theta^\circ=45^\circ

\displaystyle \text{(iii) }\sin^2\theta^\circ-\cos^2\theta^\circ
\displaystyle =\sin^2 45^\circ-\cos^2 45^\circ
\displaystyle =\left(\frac1{\sqrt2}\right)^2-\left(\frac1{\sqrt2}\right)^2
\displaystyle =\frac12-\frac12=0

\displaystyle \text{(iv) }\sin\theta^\circ=\frac{AB}{AC}=\frac5x
\displaystyle \therefore \sin45^\circ=\frac5x
\displaystyle \Rightarrow \frac1{\sqrt2}=\frac5x
\displaystyle \therefore x=5\sqrt2
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Find the magnitude of angle }A,\text{ if:}
\displaystyle \text{(i) }2\sin A\cos A-\cos A-2\sin A+1=0
\displaystyle \text{(ii) }\tan A-2\cos A\tan A+2\cos A-1=0
\displaystyle \text{(iii) }2\cos^2 A-3\cos A+1=0
\displaystyle \text{(iv) }2\tan3A\cos3A-\tan3A+1=2\cos3A
\displaystyle \text{Answer:}
\displaystyle \text{(i) }2\sin A\cos A-\cos A-2\sin A+1=0
\displaystyle \Rightarrow \cos A(2\sin A-1)-(2\sin A-1)=0
\displaystyle \Rightarrow (2\sin A-1)(\cos A-1)=0
\displaystyle \Rightarrow 2\sin A-1=0\quad\text{or}\quad\cos A-1=0
\displaystyle \Rightarrow \sin A=\frac12\quad\text{or}\quad\cos A=1
\displaystyle \Rightarrow \sin A=\sin30^\circ\quad\text{or}\quad\cos A=\cos0^\circ
\displaystyle \therefore A=30^\circ\text{ or }0^\circ

\displaystyle \text{(ii) }\tan A-2\cos A\tan A+2\cos A-1=0
\displaystyle \Rightarrow \tan A(1-2\cos A)-(1-2\cos A)=0
\displaystyle \Rightarrow (1-2\cos A)(\tan A-1)=0
\displaystyle \Rightarrow 1-2\cos A=0\quad\text{or}\quad\tan A-1=0
\displaystyle \Rightarrow \cos A=\frac12\quad\text{or}\quad\tan A=1
\displaystyle \Rightarrow \cos A=\cos60^\circ\quad\text{or}\quad\tan A=\tan45^\circ
\displaystyle \therefore A=60^\circ\text{ or }45^\circ

\displaystyle \text{(iii) }2\cos^2 A-3\cos A+1=0
\displaystyle \Rightarrow (2\cos A-1)(\cos A-1)=0
\displaystyle \Rightarrow 2\cos A-1=0\quad\text{or}\quad\cos A-1=0
\displaystyle \Rightarrow \cos A=\frac12\quad\text{or}\quad\cos A=1
\displaystyle \Rightarrow \cos A=\cos60^\circ\quad\text{or}\quad\cos A=\cos0^\circ
\displaystyle \therefore A=60^\circ\text{ or }0^\circ

\displaystyle \text{(iv) }2\tan3A\cos3A-\tan3A+1=2\cos3A
\displaystyle \Rightarrow 2\tan3A\cos3A-\tan3A-2\cos3A+1=0
\displaystyle \Rightarrow (\tan3A-1)(2\cos3A-1)=0
\displaystyle \Rightarrow \tan3A=1\quad\text{or}\quad\cos3A=\frac12
\displaystyle \Rightarrow \tan3A=\tan45^\circ\quad\text{or}\quad\cos3A=\cos60^\circ
\displaystyle \Rightarrow 3A=45^\circ\quad\text{or}\quad3A=60^\circ
\displaystyle \therefore A=15^\circ\text{ or }20^\circ
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Solve for }x:
\displaystyle \text{(i) }2\cos3x-1=0\qquad\text{(ii) }\cos\frac{x}{3}-1=0
\displaystyle \text{(iii) }\sin(x+10^\circ)=\frac12\qquad\text{(iv) }\cos(2x-30^\circ)=0
\displaystyle \text{(v) }2\cos(3x-15^\circ)=1\qquad\text{(vi) }\tan^2(x-5^\circ)=3
\displaystyle \text{(vii) }3\tan^2(2x-20^\circ)=1
\displaystyle \text{(viii) }\cos\left(\frac{x}{2}+10^\circ\right)=\frac{\sqrt3}{2}
\displaystyle \text{(ix) }\sin^2 x+\sin^2 30^\circ=1
\displaystyle \text{(x) }\cos^2 30^\circ+\cos^2 x=1
\displaystyle \text{(xi) }\cos^2 30^\circ+\sin^2 2x=1
\displaystyle \text{(xii) }\sin^2 60^\circ+\cos^2(3x-9^\circ)=1
\displaystyle \text{Answer:}
\displaystyle \text{(i) }2\cos3x-1=0
\displaystyle \Rightarrow \cos3x=\frac12=\cos60^\circ
\displaystyle \Rightarrow 3x=60^\circ
\displaystyle \therefore x=20^\circ

\displaystyle \text{(ii) }\cos\frac{x}{3}-1=0
\displaystyle \Rightarrow \cos\frac{x}{3}=1=\cos0^\circ
\displaystyle \Rightarrow \frac{x}{3}=0^\circ
\displaystyle \therefore x=0^\circ

\displaystyle \text{(iii) }\sin(x+10^\circ)=\frac12
\displaystyle \Rightarrow \sin(x+10^\circ)=\sin30^\circ
\displaystyle \Rightarrow x+10^\circ=30^\circ
\displaystyle \therefore x=20^\circ

\displaystyle \text{(iv) }\cos(2x-30^\circ)=0
\displaystyle \Rightarrow \cos(2x-30^\circ)=\cos90^\circ
\displaystyle \Rightarrow 2x-30^\circ=90^\circ
\displaystyle \therefore x=60^\circ

\displaystyle \text{(v) }2\cos(3x-15^\circ)=1
\displaystyle \Rightarrow \cos(3x-15^\circ)=\frac12=\cos60^\circ
\displaystyle \Rightarrow 3x-15^\circ=60^\circ
\displaystyle \Rightarrow 3x=75^\circ
\displaystyle \therefore x=25^\circ

\displaystyle \text{(vi) }\tan^2(x-5^\circ)=3
\displaystyle \Rightarrow \tan(x-5^\circ)=\sqrt3=\tan60^\circ
\displaystyle \Rightarrow x-5^\circ=60^\circ
\displaystyle \therefore x=65^\circ

\displaystyle \text{(vii) }3\tan^2(2x-20^\circ)=1
\displaystyle \Rightarrow \tan^2(2x-20^\circ)=\frac13
\displaystyle \Rightarrow \tan(2x-20^\circ)=\frac1{\sqrt3}=\tan30^\circ
\displaystyle \Rightarrow 2x-20^\circ=30^\circ
\displaystyle \Rightarrow 2x=50^\circ
\displaystyle \therefore x=25^\circ

\displaystyle \text{(viii) }\cos\left(\frac{x}{2}+10^\circ\right)=\frac{\sqrt3}{2}
\displaystyle \Rightarrow \cos\left(\frac{x}{2}+10^\circ\right)=\cos30^\circ
\displaystyle \Rightarrow \frac{x}{2}+10^\circ=30^\circ
\displaystyle \Rightarrow \frac{x}{2}=20^\circ
\displaystyle \therefore x=40^\circ

\displaystyle \text{(ix) }\sin^2 x+\sin^2 30^\circ=1
\displaystyle \Rightarrow \sin^2 x+\left(\frac12\right)^2=1
\displaystyle \Rightarrow \sin^2 x=\frac34
\displaystyle \Rightarrow \sin x=\frac{\sqrt3}{2}=\sin60^\circ
\displaystyle \therefore x=60^\circ

\displaystyle \text{(x) }\cos^2 30^\circ+\cos^2 x=1
\displaystyle \Rightarrow \left(\frac{\sqrt3}{2}\right)^2+\cos^2 x=1
\displaystyle \Rightarrow \cos^2 x=\frac14
\displaystyle \Rightarrow \cos x=\frac12=\cos60^\circ
\displaystyle \therefore x=60^\circ

\displaystyle \text{(xi) }\cos^2 30^\circ+\sin^2 2x=1
\displaystyle \Rightarrow \frac34+\sin^2 2x=1
\displaystyle \Rightarrow \sin^2 2x=\frac14
\displaystyle \Rightarrow \sin2x=\frac12=\sin30^\circ
\displaystyle \Rightarrow 2x=30^\circ
\displaystyle \therefore x=15^\circ

\displaystyle \text{(xii) }\sin^2 60^\circ+\cos^2(3x-9^\circ)=1
\displaystyle \Rightarrow \frac34+\cos^2(3x-9^\circ)=1
\displaystyle \Rightarrow \cos^2(3x-9^\circ)=\frac14
\displaystyle \Rightarrow \cos(3x-9^\circ)=\frac12=\cos60^\circ
\displaystyle \Rightarrow 3x-9^\circ=60^\circ
\displaystyle \Rightarrow 3x=69^\circ
\displaystyle \therefore x=23^\circ
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }4\cos^2 x=3\text{ and }x\text{ is an acute angle, find the value of:}
\displaystyle \text{(i) }x\qquad\text{(ii) }\cos^2 x+\cot^2 x
\displaystyle \text{(iii) }\cos3x\qquad\text{(iv) }\sin2x
\displaystyle \text{Answer:}
\displaystyle 4\cos^2 x=3
\displaystyle \Rightarrow \cos^2 x=\frac34
\displaystyle \Rightarrow \cos x=\frac{\sqrt3}{2}
\displaystyle \text{Since }x\text{ is acute and }\cos30^\circ=\frac{\sqrt3}{2},
\displaystyle \text{(i) }\therefore x=30^\circ

\displaystyle \text{(ii) }\cos^2 x+\cot^2 x
\displaystyle =\cos^2 30^\circ+\cot^2 30^\circ
\displaystyle =\left(\frac{\sqrt3}{2}\right)^2+(\sqrt3)^2
\displaystyle =\frac34+3=\frac{15}{4}

\displaystyle \text{(iii) }\cos3x=\cos(3\times30^\circ)
\displaystyle =\cos90^\circ=0

\displaystyle \text{(iv) }\sin2x=\sin(2\times30^\circ)
\displaystyle =\sin60^\circ=\frac{\sqrt3}{2}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{In }\triangle ABC,\ \angle B=90^\circ,\ AB=y\text{ units},\ BC=\sqrt3\text{ units},
\displaystyle AC=2\text{ units and angle }A=x^\circ,\text{ find:}
\displaystyle \text{(i) }\sin x^\circ\qquad\text{(ii) }x^\circ\qquad\text{(iii) }\tan x^\circ
\displaystyle \text{(iv) Use }\cos x^\circ\text{ to find the value of }y.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\sin x^\circ=\frac{BC}{AC}=\frac{\sqrt3}{2}

\displaystyle \text{(ii) Since }\sin x^\circ=\frac{\sqrt3}{2}=\sin60^\circ,
\displaystyle \therefore x^\circ=60^\circ

\displaystyle \text{(iii) }\tan x^\circ=\tan60^\circ=\sqrt3

\displaystyle \text{(iv) }\cos x^\circ=\frac{AB}{AC}=\frac{y}{2}
\displaystyle \therefore \cos60^\circ=\frac{y}{2}
\displaystyle \Rightarrow \frac12=\frac{y}{2}
\displaystyle \therefore y=1\text{ unit}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }2\cos(A+B)=2\sin(A-B)=1,\text{ find the values of }A\text{ and }B.
\displaystyle \text{Answer:}
\displaystyle 2\cos(A+B)=1
\displaystyle \Rightarrow \cos(A+B)=\frac12=\cos60^\circ
\displaystyle \Rightarrow A+B=60^\circ\qquad\ldots\text{(i)}
\displaystyle 2\sin(A-B)=1
\displaystyle \Rightarrow \sin(A-B)=\frac12=\sin30^\circ
\displaystyle \Rightarrow A-B=30^\circ\qquad\ldots\text{(ii)}
\displaystyle \text{Adding (i) and (ii),}
\displaystyle 2A=90^\circ
\displaystyle \therefore A=45^\circ
\displaystyle \text{Substituting }A=45^\circ\text{ in (i),}
\displaystyle 45^\circ+B=60^\circ
\displaystyle \therefore B=15^\circ
\displaystyle \therefore A=45^\circ\text{ and }B=15^\circ
\displaystyle \\

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