\displaystyle \textbf{Question 1: }\text{Evaluate each of the following:}
\displaystyle \text{(i) }\mathrm{cosec}\,30^\circ+\cot45^\circ
\displaystyle \text{(ii) }\cos30^\circ\cos45^\circ-\sin30^\circ\sin45^\circ
\displaystyle \text{(iii) }\tan30^\circ\sec45^\circ+\tan60^\circ\sec30^\circ
\displaystyle \text{(iv) }\sin30^\circ\cos45^\circ+\cos30^\circ\sin45^\circ
\displaystyle \text{(v) }\frac{\sin^2 45^\circ+\cos^2 45^\circ}{\tan^2 60^\circ}
\displaystyle \text{(vi) }\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}
\displaystyle \text{(vii) }\frac{\sin60^\circ}{\cos^2 45^\circ}-\cot30^\circ+15\cos90^\circ
\displaystyle \text{(viii) }\frac{5\sin^2 30^\circ+\cos^2 45^\circ-4\tan^2 30^\circ}{2\sin30^\circ\cos30^\circ+\tan45^\circ}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\mathrm{cosec}\,30^\circ+\cot45^\circ
\displaystyle =2+1=3

\displaystyle \text{(ii) }\cos30^\circ\cos45^\circ-\sin30^\circ\sin45^\circ
\displaystyle =\frac{\sqrt3}{2}\times\frac{1}{\sqrt2}-\frac12\times\frac{1}{\sqrt2}
\displaystyle =\frac{\sqrt3-1}{2\sqrt2}

\displaystyle \text{(iii) }\tan30^\circ\sec45^\circ+\tan60^\circ\sec30^\circ
\displaystyle =\frac{1}{\sqrt3}\times\sqrt2+\sqrt3\times\frac{2}{\sqrt3}
\displaystyle =\frac{\sqrt2}{\sqrt3}+2
\displaystyle =\frac{\sqrt2+2\sqrt3}{\sqrt3}

\displaystyle \text{(iv) }\sin30^\circ\cos45^\circ+\cos30^\circ\sin45^\circ
\displaystyle =\frac12\times\frac1{\sqrt2}+\frac{\sqrt3}{2}\times\frac1{\sqrt2}
\displaystyle =\frac{\sqrt3+1}{2\sqrt2}

\displaystyle \text{(v) }\frac{\sin^2 45^\circ+\cos^2 45^\circ}{\tan^2 60^\circ}
\displaystyle =\frac{\left(\frac1{\sqrt2}\right)^2+\left(\frac1{\sqrt2}\right)^2}{(\sqrt3)^2}
\displaystyle =\frac{\frac12+\frac12}{3}=\frac13

\displaystyle \text{(vi) }\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}
\displaystyle =\frac{\frac12-1+2\times1}{\frac1{\sqrt3}\times\sqrt3}
\displaystyle =\frac{\frac32}{1}=\frac32

\displaystyle \text{(vii) }\frac{\sin60^\circ}{\cos^2 45^\circ}-\cot30^\circ+15\cos90^\circ
\displaystyle =\frac{\frac{\sqrt3}{2}}{\left(\frac1{\sqrt2}\right)^2}-\sqrt3+15\times0
\displaystyle =\frac{\frac{\sqrt3}{2}}{\frac12}-\sqrt3
\displaystyle =\sqrt3-\sqrt3=0

\displaystyle \text{(viii) }\frac{5\sin^2 30^\circ+\cos^2 45^\circ-4\tan^2 30^\circ}{2\sin30^\circ\cos30^\circ+\tan45^\circ}
\displaystyle =\frac{5\left(\frac12\right)^2+\left(\frac1{\sqrt2}\right)^2-4\left(\frac1{\sqrt3}\right)^2}{2\times\frac12\times\frac{\sqrt3}{2}+1}
\displaystyle =\frac{\frac54+\frac12-\frac43}{\frac{\sqrt3}{2}+1}
\displaystyle =\frac{\frac5{12}}{\frac{\sqrt3+2}{2}}
\displaystyle =\frac{5}{6(\sqrt3+2)}
\displaystyle =\frac56(2-\sqrt3)
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the value of }\theta\text{ in each of the following when }0^\circ<\theta<90^\circ:
\displaystyle \text{(i) }2\sin2\theta=\sqrt3
\displaystyle \text{(ii) }2\cos3\theta=1
\displaystyle \text{(iii) }\sqrt3\tan2\theta-3=0
\displaystyle \text{(iv) }2\cos\theta=1
\displaystyle \text{(v) }2\cos^2\theta=\frac12
\displaystyle \text{(vi) }2\sin^2\theta=\frac12
\displaystyle \text{(vii) }3\tan^2\theta-1=0
\displaystyle \text{Answer:}
\displaystyle \text{Given, }0^\circ<\theta<90^\circ.
\displaystyle \text{(i) }2\sin2\theta=\sqrt3
\displaystyle \Rightarrow \sin2\theta=\frac{\sqrt3}{2}
\displaystyle \Rightarrow \sin2\theta=\sin60^\circ
\displaystyle \text{Since }0^\circ<2\theta<180^\circ,
\displaystyle 2\theta=60^\circ\text{ or }120^\circ
\displaystyle \therefore \theta=30^\circ\text{ or }60^\circ

\displaystyle \text{(ii) }2\cos3\theta=1
\displaystyle \Rightarrow \cos3\theta=\frac12=\cos60^\circ
\displaystyle \Rightarrow 3\theta=60^\circ
\displaystyle \therefore \theta=20^\circ

\displaystyle \text{(iii) }\sqrt3\tan2\theta-3=0
\displaystyle \Rightarrow \sqrt3\tan2\theta=3
\displaystyle \Rightarrow \tan2\theta=\sqrt3=\tan60^\circ
\displaystyle \Rightarrow 2\theta=60^\circ
\displaystyle \therefore \theta=30^\circ

\displaystyle \text{(iv) }2\cos\theta=1
\displaystyle \Rightarrow \cos\theta=\frac12=\cos60^\circ
\displaystyle \therefore \theta=60^\circ

\displaystyle \text{(v) }2\cos^2\theta=\frac12
\displaystyle \Rightarrow \cos^2\theta=\frac14
\displaystyle \Rightarrow \cos\theta=\frac12
\displaystyle \text{Since }\theta\text{ is acute, }\cos\theta>0.
\displaystyle \Rightarrow \cos\theta=\cos60^\circ
\displaystyle \therefore \theta=60^\circ

\displaystyle \text{(vi) }2\sin^2\theta=\frac12
\displaystyle \Rightarrow \sin^2\theta=\frac14
\displaystyle \Rightarrow \sin\theta=\frac12
\displaystyle \text{Since }\theta\text{ is acute, }\sin\theta>0.
\displaystyle \Rightarrow \sin\theta=\sin30^\circ
\displaystyle \therefore \theta=30^\circ

\displaystyle \text{(vii) }3\tan^2\theta-1=0
\displaystyle \Rightarrow \tan^2\theta=\frac13
\displaystyle \Rightarrow \tan\theta=\frac1{\sqrt3}
\displaystyle \text{Since }\theta\text{ is acute, }\tan\theta>0.
\displaystyle \Rightarrow \tan\theta=\tan30^\circ
\displaystyle \therefore \theta=30^\circ
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }\theta\text{ is an acute angle and }\tan\theta+\cot\theta=2,\text{ find the value of}
\displaystyle \tan^7\theta+\cot^7\theta.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\tan\theta+\cot\theta=2
\displaystyle \Rightarrow \tan\theta+\frac{1}{\tan\theta}=2
\displaystyle \Rightarrow \tan^2\theta+1=2\tan\theta
\displaystyle \Rightarrow \tan^2\theta-2\tan\theta+1=0
\displaystyle \Rightarrow (\tan\theta-1)^2=0
\displaystyle \Rightarrow \tan\theta=1
\displaystyle \therefore \cot\theta=1
\displaystyle \therefore \tan^7\theta+\cot^7\theta
\displaystyle =1^7+1^7=2
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }x=30^\circ,\text{ verify that:}
\displaystyle \text{(i) }\sin3x=3\sin x-4\sin^3 x
\displaystyle \text{(ii) }\cos3x=4\cos^3 x-3\cos x
\displaystyle \text{(iii) }\tan2x=\frac{2\tan x}{1-\tan^2 x}
\displaystyle \text{(iv) }\sin x=\sqrt{\frac{1-\cos2x}{2}}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x=30^\circ.
\displaystyle \text{(i) L.H.S.}=\sin3x=\sin90^\circ=1
\displaystyle \text{R.H.S.}=3\sin x-4\sin^3 x
\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 verified.}

\displaystyle \text{(ii) L.H.S.}=\cos3x=\cos90^\circ=0
\displaystyle \text{R.H.S.}=4\cos^3 x-3\cos x
\displaystyle =4\left(\frac{\sqrt3}{2}\right)^3-3\left(\frac{\sqrt3}{2}\right)
\displaystyle =\frac{3\sqrt3}{2}-\frac{3\sqrt3}{2}=0
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence verified.}

\displaystyle \text{(iii) L.H.S.}=\tan2x=\tan60^\circ=\sqrt3
\displaystyle \text{R.H.S.}=\frac{2\tan x}{1-\tan^2 x}
\displaystyle =\frac{2\tan30^\circ}{1-\tan^2 30^\circ}
\displaystyle =\frac{2\times\frac1{\sqrt3}}{1-\frac13}
\displaystyle =\frac{\frac2{\sqrt3}}{\frac23}=\sqrt3
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence verified.}

\displaystyle \text{(iv) L.H.S.}=\sin x=\sin30^\circ=\frac12
\displaystyle \text{R.H.S.}=\sqrt{\frac{1-\cos2x}{2}}
\displaystyle =\sqrt{\frac{1-\cos60^\circ}{2}}
\displaystyle =\sqrt{\frac{1-\frac12}{2}}
\displaystyle =\sqrt{\frac14}=\frac12
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence verified.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find an acute angle }\theta\text{ when }\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}
\displaystyle =\frac{1-\sqrt3}{1+\sqrt3}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}=\frac{1-\sqrt3}{1+\sqrt3}
\displaystyle \text{Applying componendo and dividendo,}
\displaystyle \frac{(\cos\theta-\sin\theta)+(\cos\theta+\sin\theta)}{(\cos\theta-\sin\theta)-(\cos\theta+\sin\theta)}
\displaystyle =\frac{(1-\sqrt3)+(1+\sqrt3)}{(1-\sqrt3)-(1+\sqrt3)}
\displaystyle \Rightarrow \frac{2\cos\theta}{-2\sin\theta}=\frac{2}{-2\sqrt3}
\displaystyle \Rightarrow -\cot\theta=-\frac{1}{\sqrt3}
\displaystyle \Rightarrow \cot\theta=\frac{1}{\sqrt3}
\displaystyle \Rightarrow \tan\theta=\sqrt3=\tan60^\circ
\displaystyle \therefore \theta=60^\circ
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }\sin(A+B)=1\text{ and }\cos(A-B)=\frac{\sqrt3}{2},\ 0^\circ<A+B\leq90^\circ,
\displaystyle A>B,\text{ then find }A\text{ and }B.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\sin(A+B)=1
\displaystyle \Rightarrow \sin(A+B)=\sin90^\circ
\displaystyle \text{Since }0^\circ<A+B\leq90^\circ,
\displaystyle A+B=90^\circ\qquad\ldots\text{(i)}
\displaystyle \text{Also, }\cos(A-B)=\frac{\sqrt3}{2}
\displaystyle \Rightarrow \cos(A-B)=\cos30^\circ
\displaystyle \text{Since }A>B\text{ and }A+B\leq90^\circ,\quad0^\circ<A-B<90^\circ.
\displaystyle \therefore A-B=30^\circ\qquad\ldots\text{(ii)}
\displaystyle \text{Adding (i) and (ii),}
\displaystyle 2A=120^\circ
\displaystyle \therefore A=60^\circ
\displaystyle \text{Substituting }A=60^\circ\text{ in (i),}
\displaystyle 60^\circ+B=90^\circ
\displaystyle \therefore B=30^\circ
\displaystyle \therefore A=60^\circ\text{ and }B=30^\circ
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{ABC is a right triangle, right-angled at }C.\text{ If }A=30^\circ\text{ and }AB=40\text{ units,}
\displaystyle \text{find the remaining two sides and }\angle B\text{ in }\triangle ABC.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\angle A=30^\circ,\quad \angle C=90^\circ\text{ and }AB=40\text{ units}.
\displaystyle \text{Since }\angle A+\angle B+\angle C=180^\circ,
\displaystyle 30^\circ+\angle B+90^\circ=180^\circ
\displaystyle \therefore \angle B=60^\circ
\displaystyle \sin30^\circ=\frac{BC}{AB}
\displaystyle \Rightarrow \frac12=\frac{BC}{40}
\displaystyle \therefore BC=40\times\frac12=20\text{ units}
\displaystyle \cos30^\circ=\frac{AC}{AB}
\displaystyle \Rightarrow \frac{\sqrt3}{2}=\frac{AC}{40}
\displaystyle \therefore AC=40\times\frac{\sqrt3}{2}=20\sqrt3\text{ units}
\displaystyle \therefore AC=20\sqrt3\text{ units},\quad BC=20\text{ units and }\angle B=60^\circ
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{A rhombus of side }20\text{ cm has two angles of }60^\circ\text{ each. Find the length}
\displaystyle \text{of the diagonals.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }ABCD\text{ is a rhombus of side }20\text{ cm}.
\displaystyle \therefore AB=BC=CD=DA=20\text{ cm}
\displaystyle \text{The diagonals of a rhombus bisect each other at right angles and bisect the vertex angles.}
\displaystyle \therefore \angle BAO=30^\circ
\displaystyle \text{In right-angled }\triangle AOB,
\displaystyle \cos30^\circ=\frac{OA}{AB}
\displaystyle \Rightarrow \frac{\sqrt3}{2}=\frac{OA}{20}
\displaystyle \therefore OA=20\times\frac{\sqrt3}{2}=10\sqrt3\text{ cm}
\displaystyle \sin30^\circ=\frac{OB}{AB}
\displaystyle \Rightarrow \frac12=\frac{OB}{20}
\displaystyle \therefore OB=10\text{ cm}
\displaystyle \text{Since the diagonals bisect each other,}
\displaystyle AC=2OA=2\times10\sqrt3=20\sqrt3\text{ cm}
\displaystyle BD=2OB=2\times10=20\text{ cm}
\displaystyle \therefore AC=20\sqrt3\text{ cm and }BD=20\text{ cm}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{An equilateral triangle is inscribed in a circle of radius }6\text{ cm. Find the side.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }OA=OB=OC=6\text{ cm}.
\displaystyle \text{Since }ABC\text{ is equilateral, }\angle A=\angle B=\angle C=60^\circ.
\displaystyle \text{Also, }OD\perp AB\text{ and }D\text{ is the midpoint of }AB.
\displaystyle \therefore \angle OBD=30^\circ
\displaystyle \text{In right-angled }\triangle OBD,
\displaystyle \cos30^\circ=\frac{BD}{OB}
\displaystyle \Rightarrow \frac{\sqrt3}{2}=\frac{BD}{6}
\displaystyle \therefore BD=6\times\frac{\sqrt3}{2}=3\sqrt3\text{ cm}
\displaystyle \text{Since }D\text{ is the midpoint of }AB,
\displaystyle AB=2BD=2\times3\sqrt3
\displaystyle \therefore AB=6\sqrt3\text{ cm}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If each of }\alpha,\beta\text{ and }\gamma\text{ are positive acute angles such that}
\displaystyle \sin(\alpha+\beta-\gamma)=\frac12,\quad \cos(\beta+\gamma-\alpha)=\frac12\text{ and}
\displaystyle \tan(\gamma+\alpha-\beta)=1,\text{ find the values of }\alpha,\beta\text{ and }\gamma.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\sin(\alpha+\beta-\gamma)=\frac12
\displaystyle \Rightarrow \sin(\alpha+\beta-\gamma)=\sin30^\circ
\displaystyle \therefore \alpha+\beta-\gamma=30^\circ\qquad\ldots\text{(i)}
\displaystyle \text{Also, }\cos(\beta+\gamma-\alpha)=\frac12
\displaystyle \Rightarrow \cos(\beta+\gamma-\alpha)=\cos60^\circ
\displaystyle \therefore \beta+\gamma-\alpha=60^\circ\qquad\ldots\text{(ii)}
\displaystyle \text{And, }\tan(\gamma+\alpha-\beta)=1
\displaystyle \Rightarrow \tan(\gamma+\alpha-\beta)=\tan45^\circ
\displaystyle \therefore \gamma+\alpha-\beta=45^\circ\qquad\ldots\text{(iii)}
\displaystyle \text{Adding (i), (ii) and (iii),}
\displaystyle \alpha+\beta+\gamma=135^\circ\qquad\ldots\text{(iv)}
\displaystyle \text{Subtracting (i) from (iv),}
\displaystyle 2\gamma=105^\circ
\displaystyle \therefore \gamma=52\frac12^\circ
\displaystyle \text{Subtracting (ii) from (iv),}
\displaystyle 2\alpha=75^\circ
\displaystyle \therefore \alpha=37\frac12^\circ
\displaystyle \text{Subtracting (iii) from (iv),}
\displaystyle 2\beta=90^\circ
\displaystyle \therefore \beta=45^\circ
\displaystyle \therefore \alpha=37\frac12^\circ,\quad \beta=45^\circ\text{ and }\gamma=52\frac12^\circ
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In an acute-angled triangle }ABC,\text{ if }\tan(A+B-C)=1\text{ and}
\displaystyle \sec(B+C-A)=2,\text{ find the values of }A,\ B\text{ and }C.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\tan(A+B-C)=1
\displaystyle \Rightarrow \tan(A+B-C)=\tan45^\circ
\displaystyle \therefore A+B-C=45^\circ\qquad\ldots\text{(i)}
\displaystyle \text{Also, }\sec(B+C-A)=2
\displaystyle \Rightarrow \sec(B+C-A)=\sec60^\circ
\displaystyle \therefore B+C-A=60^\circ\qquad\ldots\text{(ii)}
\displaystyle \text{Adding (i) and (ii),}
\displaystyle 2B=105^\circ
\displaystyle \therefore B=52\frac12^\circ
\displaystyle \text{Substituting }B=52\frac12^\circ\text{ in (ii),}
\displaystyle 52\frac12^\circ+C-A=60^\circ
\displaystyle \Rightarrow C-A=7\frac12^\circ\qquad\ldots\text{(iii)}
\displaystyle \text{Since }A+B+C=180^\circ,
\displaystyle A+52\frac12^\circ+C=180^\circ
\displaystyle \Rightarrow A+C=127\frac12^\circ\qquad\ldots\text{(iv)}
\displaystyle \text{Adding (iii) and (iv),}
\displaystyle 2C=135^\circ
\displaystyle \therefore C=67\frac12^\circ
\displaystyle \text{From (iv), }A=127\frac12^\circ-67\frac12^\circ=60^\circ
\displaystyle \therefore A=60^\circ,\quad B=52\frac12^\circ\text{ and }C=67\frac12^\circ
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Evaluate:}
\displaystyle \sin^2 30^\circ\cos^2 45^\circ+4\tan^2 30^\circ+\frac12\sin^2 90^\circ-2\cos^2 90^\circ
\displaystyle +\frac1{24}\cos^2 0^\circ
\displaystyle \text{Answer:}
\displaystyle \sin^2 30^\circ\cos^2 45^\circ+4\tan^2 30^\circ+\frac12\sin^2 90^\circ-2\cos^2 90^\circ+\frac1{24}\cos^2 0^\circ
\displaystyle =\left(\frac12\right)^2\left(\frac1{\sqrt2}\right)^2+4\left(\frac1{\sqrt3}\right)^2+\frac12(1)^2-2(0)^2+\frac1{24}(1)^2
\displaystyle =\frac14\times\frac12+\frac43+\frac12+\frac1{24}
\displaystyle =\frac18+\frac43+\frac12+\frac1{24}
\displaystyle =\frac{3+32+12+1}{24}
\displaystyle =\frac{48}{24}=2
\displaystyle \therefore \text{The value is }2.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Evaluate: }\cot^2 30^\circ-2\cos^2 60^\circ-\frac34\sec^2 45^\circ-4\sec^2 30^\circ
\displaystyle \text{Answer:}
\displaystyle \cot^2 30^\circ-2\cos^2 60^\circ-\frac34\sec^2 45^\circ-4\sec^2 30^\circ
\displaystyle =(\sqrt3)^2-2\left(\frac12\right)^2-\frac34(\sqrt2)^2-4\left(\frac2{\sqrt3}\right)^2
\displaystyle =3-\frac12-\frac32-\frac{16}{3}
\displaystyle =\frac{18-3-9-32}{6}
\displaystyle =-\frac{26}{6}=-\frac{13}{3}
\displaystyle \therefore \text{The value is }-\frac{13}{3}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Find the value of }x:
\displaystyle \text{(i) }2\sin3x=\sqrt3\qquad\text{(ii) }\sqrt3\sin x=\cos x
\displaystyle \text{Answer:}
\displaystyle \text{(i) }2\sin3x=\sqrt3
\displaystyle \Rightarrow \sin3x=\frac{\sqrt3}{2}
\displaystyle \Rightarrow \sin3x=\sin60^\circ
\displaystyle \Rightarrow 3x=60^\circ
\displaystyle \therefore x=20^\circ

\displaystyle \text{(ii) }\sqrt3\sin x=\cos x
\displaystyle \Rightarrow \frac{\sin x}{\cos x}=\frac1{\sqrt3}
\displaystyle \Rightarrow \tan x=\frac1{\sqrt3}
\displaystyle \Rightarrow \tan x=\tan30^\circ
\displaystyle \therefore x=30^\circ
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }A=B=60^\circ,\text{ verify:}
\displaystyle \text{(i) }\cos(A-B)=\cos A\cos B+\sin A\sin B
\displaystyle \text{(ii) }\sin(A-B)=\sin A\cos B-\cos A\sin B
\displaystyle \text{(iii) }\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A=B=60^\circ.
\displaystyle \therefore \sin A=\sin B=\frac{\sqrt3}{2},\quad \cos A=\cos B=\frac12,\quad \tan A=\tan B=\sqrt3
\displaystyle \text{(i) L.H.S.}=\cos(A-B)
\displaystyle =\cos(60^\circ-60^\circ)=\cos0^\circ=1
\displaystyle \text{R.H.S.}=\cos A\cos B+\sin A\sin B
\displaystyle =\frac12\times\frac12+\frac{\sqrt3}{2}\times\frac{\sqrt3}{2}
\displaystyle =\frac14+\frac34=1
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence verified.}

\displaystyle \text{(ii) L.H.S.}=\sin(A-B)
\displaystyle =\sin(60^\circ-60^\circ)=\sin0^\circ=0
\displaystyle \text{R.H.S.}=\sin A\cos B-\cos A\sin B
\displaystyle =\frac{\sqrt3}{2}\times\frac12-\frac12\times\frac{\sqrt3}{2}
\displaystyle =\frac{\sqrt3}{4}-\frac{\sqrt3}{4}=0
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence verified.}

\displaystyle \text{(iii) L.H.S.}=\tan(A-B)
\displaystyle =\tan(60^\circ-60^\circ)=\tan0^\circ=0
\displaystyle \text{R.H.S.}=\frac{\tan A-\tan B}{1+\tan A\tan B}
\displaystyle =\frac{\sqrt3-\sqrt3}{1+\sqrt3\times\sqrt3}
\displaystyle =\frac{0}{4}=0
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S. Hence verified.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If }\sin(A-B)=\sin A\cos B-\cos A\sin B\text{ and}
\displaystyle \cos(A-B)=\cos A\cos B+\sin A\sin B,\text{ find the values of }\sin15^\circ\text{ and }\cos15^\circ.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\sin15^\circ=\sin(45^\circ-30^\circ)
\displaystyle =\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ
\displaystyle =\frac{1}{\sqrt2}\times\frac{\sqrt3}{2}-\frac{1}{\sqrt2}\times\frac12
\displaystyle =\frac{\sqrt3-1}{2\sqrt2}

\displaystyle \text{(ii) }\cos15^\circ=\cos(45^\circ-30^\circ)
\displaystyle =\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ
\displaystyle =\frac{1}{\sqrt2}\times\frac{\sqrt3}{2}+\frac{1}{\sqrt2}\times\frac12
\displaystyle =\frac{\sqrt3+1}{2\sqrt2}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{In a right-angled triangle, right-angled at }C,\text{ if }\angle B=60^\circ\text{ and}
\displaystyle AB=15\text{ units, find the remaining angles and sides.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\angle C=90^\circ,\quad \angle B=60^\circ\text{ and }AB=15\text{ units}.
\displaystyle \text{Since }\angle A+\angle B+\angle C=180^\circ,
\displaystyle \angle A=180^\circ-90^\circ-60^\circ=30^\circ
\displaystyle \sin30^\circ=\frac{BC}{AB}
\displaystyle \Rightarrow \frac12=\frac{BC}{15}
\displaystyle \therefore BC=\frac{15}{2}=7\frac12\text{ units}
\displaystyle \cos30^\circ=\frac{AC}{AB}
\displaystyle \Rightarrow \frac{\sqrt3}{2}=\frac{AC}{15}
\displaystyle \therefore AC=\frac{15\sqrt3}{2}=7\frac12\sqrt3\text{ units}
\displaystyle \therefore \angle A=30^\circ,\quad BC=7\frac12\text{ units and }AC=7\frac12\sqrt3\text{ units}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If }\sin(A+B)=1\text{ and }\cos(A-B)=1,\ 0^\circ<A+B\leq90^\circ
\displaystyle \text{and }A\geq B,\text{ find }A\text{ and }B.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\sin(A+B)=1
\displaystyle \Rightarrow \sin(A+B)=\sin90^\circ
\displaystyle \text{Since }0^\circ<A+B\leq90^\circ,
\displaystyle A+B=90^\circ\qquad\ldots\text{(i)}
\displaystyle \text{Also, }\cos(A-B)=1
\displaystyle \Rightarrow \cos(A-B)=\cos0^\circ
\displaystyle \text{Since }A\geq B,\quad A-B\geq0^\circ.
\displaystyle \therefore A-B=0^\circ
\displaystyle \therefore A=B\qquad\ldots\text{(ii)}
\displaystyle \text{Using (i) and (ii),}
\displaystyle 2A=90^\circ
\displaystyle \therefore A=45^\circ
\displaystyle \therefore B=45^\circ
\displaystyle \therefore A=45^\circ\text{ and }B=45^\circ
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If }\tan(A-B)=\frac{1}{\sqrt3}\text{ and }\tan(A+B)=\sqrt3,\ 0^\circ<A+B\leq90^\circ
\displaystyle \text{and }A>B,\text{ find }A\text{ and }B.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\tan(A-B)=\frac{1}{\sqrt3}
\displaystyle \Rightarrow \tan(A-B)=\tan30^\circ
\displaystyle \therefore A-B=30^\circ\qquad\ldots\text{(i)}
\displaystyle \text{Also, }\tan(A+B)=\sqrt3
\displaystyle \Rightarrow \tan(A+B)=\tan60^\circ
\displaystyle \text{Since }0^\circ<A+B\leq90^\circ,
\displaystyle A+B=60^\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 (ii),}
\displaystyle 45^\circ+B=60^\circ
\displaystyle \therefore B=15^\circ
\displaystyle \therefore A=45^\circ\text{ and }B=15^\circ
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{In }\triangle ABC,\text{ right-angled at }B,\ \angle A=\angle C.\text{ Find the values of:}
\displaystyle \text{(i) }\sin A\cos C+\cos A\sin C
\displaystyle \text{(ii) }\sin A\sin B+\cos A\cos B
\displaystyle \text{Answer:}
\displaystyle \text{Since }\angle A+\angle B+\angle C=180^\circ\text{ and }\angle B=90^\circ,
\displaystyle \angle A+\angle C=90^\circ
\displaystyle \text{Given, }\angle A=\angle C
\displaystyle \therefore \angle A=\angle C=45^\circ
\displaystyle \therefore \sin A=\sin C=\cos A=\cos C=\frac1{\sqrt2}
\displaystyle \text{Also, }\sin B=\sin90^\circ=1\text{ and }\cos B=\cos90^\circ=0.
\displaystyle \text{(i) }\sin A\cos C+\cos A\sin C
\displaystyle =\frac1{\sqrt2}\times\frac1{\sqrt2}+\frac1{\sqrt2}\times\frac1{\sqrt2}
\displaystyle =\frac12+\frac12=1

\displaystyle \text{(ii) }\sin A\sin B+\cos A\cos B
\displaystyle =\frac1{\sqrt2}\times1+\frac1{\sqrt2}\times0
\displaystyle =\frac1{\sqrt2}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If }A\text{ and }B\text{ are acute angles such that }\tan A=\frac12\text{ and }\tan B=\frac13,
\displaystyle \text{using }\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B},\text{ find }A+B.
\displaystyle \text{Answer:}
\displaystyle \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}
\displaystyle =\frac{\frac12+\frac13}{1-\frac12\times\frac13}
\displaystyle =\frac{\frac56}{\frac56}=1
\displaystyle \therefore \tan(A+B)=1=\tan45^\circ
\displaystyle \therefore A+B=45^\circ
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{In the adjoining figure, }ABC\text{ is a right-angled triangle at }B\text{ and }\triangle ABD
\displaystyle \text{is a right-angled triangle at }A.\text{ If }BD\perp AC\text{ and }BC=2\sqrt3,\text{ find }AD.
\displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle ABC,\quad \angle CAB=30^\circ.
\displaystyle \tan30^\circ=\frac{BC}{AB}
\displaystyle \Rightarrow \frac1{\sqrt3}=\frac{2\sqrt3}{AB}
\displaystyle \Rightarrow AB=6
\displaystyle \text{Since }BD\perp AC\text{ and }\angle CAB=30^\circ,
\displaystyle \angle DBA=90^\circ-30^\circ=60^\circ
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle \tan60^\circ=\frac{AD}{AB}
\displaystyle \Rightarrow \sqrt3=\frac{AD}{6}
\displaystyle \therefore AD=6\sqrt3
\displaystyle \\

 

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