\displaystyle \textbf{Question 1: }\text{Evaluate the following:}
\displaystyle \text{(i) }\frac{\sin20^\circ}{\cos70^\circ}\qquad\text{(ii) }\frac{\cos19^\circ}{\sin71^\circ}\qquad\text{(iii) }\frac{\sin21^\circ}{\cos69^\circ}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\frac{\sin20^\circ}{\cos70^\circ}
\displaystyle =\frac{\sin20^\circ}{\cos(90^\circ-20^\circ)}
\displaystyle =\frac{\sin20^\circ}{\sin20^\circ}=1
\displaystyle \therefore \text{The required value is }1.

\displaystyle \text{(ii) }\frac{\cos19^\circ}{\sin71^\circ}
\displaystyle =\frac{\cos19^\circ}{\sin(90^\circ-19^\circ)}
\displaystyle =\frac{\cos19^\circ}{\cos19^\circ}=1
\displaystyle \therefore \text{The required value is }1.

\displaystyle \text{(iii) }\frac{\sin21^\circ}{\cos69^\circ}
\displaystyle =\frac{\sin21^\circ}{\cos(90^\circ-21^\circ)}
\displaystyle =\frac{\sin21^\circ}{\sin21^\circ}=1
\displaystyle \therefore \text{The required value is }1.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Evaluate the following:}
\displaystyle \text{(i) }\left(\frac{\sin49^\circ}{\cos41^\circ}\right)^2+\left(\frac{\cos41^\circ}{\sin49^\circ}\right)^2
\displaystyle \text{(ii) }\frac{\cot40^\circ}{\tan50^\circ}-\frac12\left(\frac{\cos35^\circ}{\sin55^\circ}\right)
\displaystyle \text{(iii) }\frac{\tan35^\circ}{\cot55^\circ}+\frac{\cot78^\circ}{\tan12^\circ}-1
\displaystyle \text{(iv) }\frac{\sec70^\circ}{\mathrm{cosec}\,20^\circ}+\frac{\sin59^\circ}{\cos31^\circ}
\displaystyle \text{(v) }\tan48^\circ\tan23^\circ\tan42^\circ\tan67^\circ
\displaystyle \text{(vi) }\sec50^\circ\sin40^\circ+\cos40^\circ\mathrm{cosec}\,50^\circ
\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }\cos41^\circ=\sin(90^\circ-41^\circ)=\sin49^\circ,
\displaystyle \left(\frac{\sin49^\circ}{\cos41^\circ}\right)^2+\left(\frac{\cos41^\circ}{\sin49^\circ}\right)^2
\displaystyle =\left(\frac{\sin49^\circ}{\sin49^\circ}\right)^2+\left(\frac{\sin49^\circ}{\sin49^\circ}\right)^2
\displaystyle =1+1=2
\displaystyle \therefore \text{The required value is }2.

\displaystyle \text{(ii) Since }\tan50^\circ=\cot40^\circ\text{ and }\sin55^\circ=\cos35^\circ,
\displaystyle \frac{\cot40^\circ}{\tan50^\circ}-\frac12\left(\frac{\cos35^\circ}{\sin55^\circ}\right)
\displaystyle =\frac{\cot40^\circ}{\cot40^\circ}-\frac12\left(\frac{\cos35^\circ}{\cos35^\circ}\right)
\displaystyle =1-\frac12=\frac12
\displaystyle \therefore \text{The required value is }\frac12.

\displaystyle \text{(iii) Since }\cot55^\circ=\tan35^\circ\text{ and }\cot78^\circ=\tan12^\circ,
\displaystyle \frac{\tan35^\circ}{\cot55^\circ}+\frac{\cot78^\circ}{\tan12^\circ}-1
\displaystyle =\frac{\tan35^\circ}{\tan35^\circ}+\frac{\tan12^\circ}{\tan12^\circ}-1
\displaystyle =1+1-1=1
\displaystyle \therefore \text{The required value is }1.

\displaystyle \text{(iv) Since }\sec70^\circ=\mathrm{cosec}\,20^\circ\text{ and }\sin59^\circ=\cos31^\circ,
\displaystyle \frac{\sec70^\circ}{\mathrm{cosec}\,20^\circ}+\frac{\sin59^\circ}{\cos31^\circ}
\displaystyle =\frac{\mathrm{cosec}\,20^\circ}{\mathrm{cosec}\,20^\circ}+\frac{\cos31^\circ}{\cos31^\circ}
\displaystyle =1+1=2
\displaystyle \therefore \text{The required value is }2.

\displaystyle \text{(v) Since }\tan48^\circ=\cot42^\circ\text{ and }\tan23^\circ=\cot67^\circ,
\displaystyle \tan48^\circ\tan23^\circ\tan42^\circ\tan67^\circ
\displaystyle =\cot42^\circ\cot67^\circ\tan42^\circ\tan67^\circ
\displaystyle =(\cot42^\circ\tan42^\circ)(\cot67^\circ\tan67^\circ)
\displaystyle =1\times1=1
\displaystyle \therefore \text{The required value is }1.

\displaystyle \text{(vi) Since }\sec50^\circ=\frac{1}{\cos50^\circ}=\frac{1}{\sin40^\circ},
\displaystyle \text{and }\mathrm{cosec}\,50^\circ=\frac{1}{\sin50^\circ}=\frac{1}{\cos40^\circ},
\displaystyle \sec50^\circ\sin40^\circ+\cos40^\circ\mathrm{cosec}\,50^\circ
\displaystyle =\frac{\sin40^\circ}{\sin40^\circ}+\frac{\cos40^\circ}{\cos40^\circ}
\displaystyle =1+1=2
\displaystyle \therefore \text{The required value is }2.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Express each one of the following in terms of trigonometric ratios of angles lying}
\displaystyle \text{between }0^\circ\text{ and }45^\circ.
\displaystyle \text{(i) }\cos78^\circ+\sec78^\circ\qquad\text{(ii) }\mathrm{cosec}\,54^\circ+\sin72^\circ
\displaystyle \text{(iii) }\cot85^\circ+\cos75^\circ\qquad\text{(iv) }\sin67^\circ+\cos75^\circ
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\cos78^\circ+\sec78^\circ
\displaystyle =\cos(90^\circ-12^\circ)+\sec(90^\circ-12^\circ)
\displaystyle =\sin12^\circ+\mathrm{cosec}\,12^\circ
\displaystyle \therefore \text{The required expression is }\sin12^\circ+\mathrm{cosec}\,12^\circ.

\displaystyle \text{(ii) }\mathrm{cosec}\,54^\circ+\sin72^\circ
\displaystyle =\mathrm{cosec}(90^\circ-36^\circ)+\sin(90^\circ-18^\circ)
\displaystyle =\sec36^\circ+\cos18^\circ
\displaystyle \therefore \text{The required expression is }\sec36^\circ+\cos18^\circ.

\displaystyle \text{(iii) }\cot85^\circ+\cos75^\circ
\displaystyle =\cot(90^\circ-5^\circ)+\cos(90^\circ-15^\circ)
\displaystyle =\tan5^\circ+\sin15^\circ
\displaystyle \therefore \text{The required expression is }\tan5^\circ+\sin15^\circ.

\displaystyle \text{(iv) }\sin67^\circ+\cos75^\circ
\displaystyle =\sin(90^\circ-23^\circ)+\cos(90^\circ-15^\circ)
\displaystyle =\cos23^\circ+\sin15^\circ
\displaystyle \therefore \text{The required expression is }\cos23^\circ+\sin15^\circ.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Express }\cos75^\circ+\cot75^\circ\text{ in terms of angles between }0^\circ\text{ and }30^\circ.
\displaystyle \text{Answer:}
\displaystyle \cos75^\circ+\cot75^\circ
\displaystyle =\cos(90^\circ-15^\circ)+\cot(90^\circ-15^\circ)
\displaystyle =\sin15^\circ+\tan15^\circ
\displaystyle \therefore \text{The required expression is }\sin15^\circ+\tan15^\circ.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }\sin3A=\cos(A-26^\circ),\text{ where }3A\text{ is an acute angle, find the value of }A.
\displaystyle \text{Answer:}
\displaystyle \sin3A=\cos(A-26^\circ)
\displaystyle \text{Using }\sin\theta=\cos(90^\circ-\theta),
\displaystyle 3A+(A-26^\circ)=90^\circ
\displaystyle 4A-26^\circ=90^\circ
\displaystyle 4A=116^\circ
\displaystyle \therefore A=29^\circ.
\displaystyle \text{Also, }3A=87^\circ,\text{ which is an acute angle.}
\displaystyle \therefore \text{The required value of }A\text{ is }29^\circ.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }A,\ B,\ C\text{ are the interior angles of a triangle }ABC,\text{ prove that:}
\displaystyle \text{(i) }\tan\left(\frac{C+A}{2}\right)=\cot\frac{B}{2}
\displaystyle \text{(ii) }\sin\left(\frac{B+C}{2}\right)=\cos\frac{A}{2}\qquad\hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }A+B+C=180^\circ,

\displaystyle \text{(i) }A+C=180^\circ-B
\displaystyle \frac{A+C}{2}=90^\circ-\frac{B}{2}
\displaystyle \therefore \tan\left(\frac{C+A}{2}\right)=\tan\left(90^\circ-\frac{B}{2}\right)
\displaystyle =\cot\frac{B}{2}
\displaystyle \therefore \tan\left(\frac{C+A}{2}\right)=\cot\frac{B}{2}.
\displaystyle \text{Hence proved.}

\displaystyle \text{(ii) }B+C=180^\circ-A
\displaystyle \frac{B+C}{2}=90^\circ-\frac{A}{2}
\displaystyle \therefore \sin\left(\frac{B+C}{2}\right)=\sin\left(90^\circ-\frac{A}{2}\right)
\displaystyle =\cos\frac{A}{2}
\displaystyle \therefore \sin\left(\frac{B+C}{2}\right)=\cos\frac{A}{2}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{(iii) If }\angle A=90^\circ,\text{ then find the value of }\tan\left(\frac{B+C}{2}\right).
\displaystyle \hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle A+B+C=180^\circ
\displaystyle 90^\circ+B+C=180^\circ
\displaystyle \therefore B+C=90^\circ
\displaystyle \therefore \frac{B+C}{2}=45^\circ
\displaystyle \tan\left(\frac{B+C}{2}\right)=\tan45^\circ=1
\displaystyle \therefore \text{The required value is }1.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Prove the following:}
\displaystyle \text{(i) }\sin\theta\sin(90^\circ-\theta)-\cos\theta\cos(90^\circ-\theta)=0
\displaystyle \text{(ii) }\frac{\cos(90^\circ-\theta)\sec(90^\circ-\theta)\tan\theta}{\mathrm{cosec}(90^\circ-\theta)\sin(90^\circ-\theta)\cot(90^\circ-\theta)}+\frac{\tan(90^\circ-\theta)}{\cot\theta}=2
\displaystyle \text{Answer:}
\displaystyle \text{(i) LHS}=\sin\theta\sin(90^\circ-\theta)-\cos\theta\cos(90^\circ-\theta)
\displaystyle =\sin\theta\cos\theta-\cos\theta\sin\theta
\displaystyle =0=\text{RHS}
\displaystyle \therefore \sin\theta\sin(90^\circ-\theta)-\cos\theta\cos(90^\circ-\theta)=0.
\displaystyle \text{Hence proved.}

\displaystyle \text{(ii) LHS}=\frac{\cos(90^\circ-\theta)\sec(90^\circ-\theta)\tan\theta}{\mathrm{cosec}(90^\circ-\theta)\sin(90^\circ-\theta)\cot(90^\circ-\theta)}+\frac{\tan(90^\circ-\theta)}{\cot\theta}
\displaystyle =\frac{\sin\theta\,\mathrm{cosec}\,\theta\,\tan\theta}{\sec\theta\,\cos\theta\,\tan\theta}+\frac{\cot\theta}{\cot\theta}
\displaystyle =\frac{(1)\tan\theta}{(1)\tan\theta}+1
\displaystyle =1+1=2=\text{RHS}
\displaystyle \therefore \frac{\cos(90^\circ-\theta)\sec(90^\circ-\theta)\tan\theta}{\mathrm{cosec}(90^\circ-\theta)\sin(90^\circ-\theta)\cot(90^\circ-\theta)}+\frac{\tan(90^\circ-\theta)}{\cot\theta}=2.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Evaluate:}
\displaystyle \text{(i) }\tan7^\circ\tan23^\circ\tan60^\circ\tan67^\circ\tan83^\circ\hfill\text{[CBSE 2000]}
\displaystyle \text{Answer:}
\displaystyle \tan83^\circ=\cot7^\circ,\qquad\tan67^\circ=\cot23^\circ
\displaystyle \therefore \tan7^\circ\tan23^\circ\tan60^\circ\tan67^\circ\tan83^\circ
\displaystyle =\tan7^\circ\cot7^\circ\tan23^\circ\cot23^\circ\tan60^\circ
\displaystyle =1\times1\times\sqrt3=\sqrt3
\displaystyle \therefore \text{The required value is }\sqrt3.

\displaystyle \text{(ii) }\frac{2\sin68^\circ}{\cos22^\circ}-\frac{2\cot15^\circ}{5\tan75^\circ}-\frac{3\tan45^\circ\tan20^\circ\tan40^\circ\tan50^\circ\tan70^\circ}{5}   \hfill\text{[CBSE 2004]}
\displaystyle \text{Answer:}
\displaystyle \sin68^\circ=\cos22^\circ,\qquad\cot15^\circ=\tan75^\circ
\displaystyle \tan70^\circ=\cot20^\circ,\qquad\tan50^\circ=\cot40^\circ
\displaystyle \therefore \frac{2\sin68^\circ}{\cos22^\circ}-\frac{2\cot15^\circ}{5\tan75^\circ}
\displaystyle \qquad-\frac{3\tan45^\circ\tan20^\circ\tan40^\circ\tan50^\circ\tan70^\circ}{5}
\displaystyle =2-\frac25-\frac35(1)(1)(1)
\displaystyle =2-\frac25-\frac35=2-1=1
\displaystyle \therefore \text{The required value is }1.

\displaystyle \text{(iii) }\frac{\sin18^\circ}{\cos72^\circ}+\sqrt3\{\tan10^\circ\tan30^\circ\tan40^\circ\tan50^\circ\tan80^\circ\}   \hfill\text{[CBSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \sin18^\circ=\cos72^\circ,\quad\tan80^\circ=\cot10^\circ,\quad\tan50^\circ=\cot40^\circ
\displaystyle \therefore \frac{\sin18^\circ}{\cos72^\circ}+\sqrt3\{\tan10^\circ\tan30^\circ\tan40^\circ\tan50^\circ\tan80^\circ\}
\displaystyle =1+\sqrt3\left(1\times\frac{1}{\sqrt3}\times1\right)
\displaystyle =1+1=2
\displaystyle \therefore \text{The required value is }2.

\displaystyle \text{(iv) }\left(\frac{3\tan41^\circ}{\cot49^\circ}\right)^2-\left(\frac{\sin35^\circ\sec55^\circ}{\tan10^\circ\tan20^\circ\tan60^\circ\tan70^\circ\tan80^\circ}\right)^2  \hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \cot49^\circ=\tan41^\circ
\displaystyle \sin35^\circ=\cos55^\circ\quad\therefore\quad\sin35^\circ\sec55^\circ=1
\displaystyle \tan80^\circ=\cot10^\circ,\qquad\tan70^\circ=\cot20^\circ
\displaystyle \therefore \left(\frac{3\tan41^\circ}{\cot49^\circ}\right)^2-\left(\frac{\sin35^\circ\sec55^\circ}{\tan10^\circ\tan20^\circ\tan60^\circ\tan70^\circ\tan80^\circ}\right)^2
\displaystyle =\left(\frac{3\tan41^\circ}{\tan41^\circ}\right)^2-\left(\frac{1}{(1)(1)(\sqrt3)}\right)^2
\displaystyle =3^2-\left(\frac{1}{\sqrt3}\right)^2
\displaystyle =9-\frac13=\frac{27-1}{3}=\frac{26}{3}
\displaystyle \therefore \text{The required value is }\frac{26}{3}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }\sin\theta=\cos(\theta-45^\circ),\text{ where }\theta\text{ and }\theta-45^\circ\text{ are acute angles,}
\displaystyle \text{find the degree measure of }\theta.
\displaystyle \text{Answer:}
\displaystyle \sin\theta=\cos(\theta-45^\circ)
\displaystyle \text{Since }\theta\text{ and }\theta-45^\circ\text{ are acute angles, they are complementary.}
\displaystyle \therefore \theta+(\theta-45^\circ)=90^\circ
\displaystyle 2\theta-45^\circ=90^\circ
\displaystyle 2\theta=135^\circ
\displaystyle \therefore \theta=67.5^\circ=67^\circ30'.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }A,\ B,\ C\text{ are the interior angles of a }\triangle ABC,\text{ show that:}
\displaystyle \text{(i) }\tan\frac{B+C}{2}=\cot\frac{A}{2}\qquad\text{(ii) }\cos\frac{B+C}{2}=\sin\frac{A}{2}
\displaystyle \text{Answer:}
\displaystyle \text{Since }A+B+C=180^\circ,

\displaystyle \text{(i) }B+C=180^\circ-A
\displaystyle \frac{B+C}{2}=90^\circ-\frac{A}{2}
\displaystyle \therefore \tan\frac{B+C}{2}=\tan\left(90^\circ-\frac{A}{2}\right)
\displaystyle =\cot\frac{A}{2}
\displaystyle \therefore \tan\frac{B+C}{2}=\cot\frac{A}{2}.
\displaystyle \text{Hence proved.}

\displaystyle \text{(ii) }\frac{B+C}{2}=90^\circ-\frac{A}{2}
\displaystyle \therefore \cos\frac{B+C}{2}=\cos\left(90^\circ-\frac{A}{2}\right)
\displaystyle =\sin\frac{A}{2}
\displaystyle \therefore \cos\frac{B+C}{2}=\sin\frac{A}{2}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If }\sin3\theta=\cos(\theta-6^\circ),\text{ where }3\theta\text{ and }\theta-6^\circ\text{ are acute angles,}
\displaystyle \text{find the value of }\theta.
\displaystyle \text{Answer:}
\displaystyle \sin3\theta=\cos(\theta-6^\circ)
\displaystyle \text{Since }3\theta\text{ and }\theta-6^\circ\text{ are acute angles, they are complementary.}
\displaystyle \therefore 3\theta+(\theta-6^\circ)=90^\circ
\displaystyle 4\theta-6^\circ=90^\circ
\displaystyle 4\theta=96^\circ
\displaystyle \therefore \theta=24^\circ.
\displaystyle \text{Also, }3\theta=72^\circ\text{ and }\theta-6^\circ=18^\circ,\text{ which are acute angles.}
\displaystyle \therefore \text{The required value of }\theta\text{ is }24^\circ.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }\sec4A=\mathrm{cosec}(A-20^\circ),\text{ where }4A\text{ is an acute angle, find the value of }A.
\displaystyle \text{Answer:}
\displaystyle \sec4A=\mathrm{cosec}(A-20^\circ)
\displaystyle \text{Using }\sec\theta=\mathrm{cosec}(90^\circ-\theta),
\displaystyle 4A+(A-20^\circ)=90^\circ
\displaystyle 5A-20^\circ=90^\circ
\displaystyle 5A=110^\circ
\displaystyle \therefore A=22^\circ.
\displaystyle \text{Also, }4A=88^\circ,\text{ which is an acute angle.}
\displaystyle \therefore \text{The required value of }A\text{ is }22^\circ.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }\sec2A=\mathrm{cosec}(A-42^\circ),\text{ where }2A\text{ is an acute angle, find the value of }A.
\displaystyle \hfill\text{[CBSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \sec2A=\mathrm{cosec}(A-42^\circ)
\displaystyle \text{Using }\sec\theta=\mathrm{cosec}(90^\circ-\theta),
\displaystyle 2A+(A-42^\circ)=90^\circ
\displaystyle 3A-42^\circ=90^\circ
\displaystyle 3A=132^\circ
\displaystyle \therefore A=44^\circ.
\displaystyle \text{Also, }2A=88^\circ,\text{ which is an acute angle.}
\displaystyle \therefore \text{The required value of }A\text{ is }44^\circ.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }\tan2A=\cot(A-18^\circ),\text{ where }2A\text{ is an acute angle, find the value of }A.
\displaystyle \hfill\text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \tan2A=\cot(A-18^\circ)
\displaystyle \text{Using }\tan\theta=\cot(90^\circ-\theta),
\displaystyle 2A+(A-18^\circ)=90^\circ
\displaystyle 3A-18^\circ=90^\circ
\displaystyle 3A=108^\circ
\displaystyle \therefore A=36^\circ.
\displaystyle \text{Also, }2A=72^\circ,\text{ which is an acute angle.}
\displaystyle \therefore \text{The required value of }A\text{ is }36^\circ.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Prove that: } \\ \sin(50^\circ+\theta)-\cos(40^\circ-\theta)   +\tan1^\circ\tan10^\circ\tan20^\circ\tan70^\circ\tan80^\circ\tan89^\circ=1.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sin(50^\circ+\theta)-\cos(40^\circ-\theta)
\displaystyle \qquad+\tan1^\circ\tan10^\circ\tan20^\circ\tan70^\circ\tan80^\circ\tan89^\circ
\displaystyle \text{Since }50^\circ+\theta=90^\circ-(40^\circ-\theta),
\displaystyle \sin(50^\circ+\theta)=\cos(40^\circ-\theta)
\displaystyle \therefore \sin(50^\circ+\theta)-\cos(40^\circ-\theta)=0
\displaystyle \text{Also, }\tan89^\circ=\cot1^\circ,\quad\tan80^\circ=\cot10^\circ,\quad\tan70^\circ=\cot20^\circ
\displaystyle \therefore \tan1^\circ\tan10^\circ\tan20^\circ\tan70^\circ\tan80^\circ\tan89^\circ
\displaystyle =(\tan1^\circ\cot1^\circ)(\tan10^\circ\cot10^\circ)(\tan20^\circ\cot20^\circ)
\displaystyle =1\times1\times1=1
\displaystyle \therefore \text{LHS}=0+1=1=\text{RHS}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Evaluate:}
\displaystyle \text{(i) }\frac{2}{3}(\cos^4 30^\circ-\sin^4 45^\circ)-3(\sin^2 60^\circ-\sec^2 45^\circ)+\frac14\cot^2 30^\circ   \hfill\text{[CBSE 2001 C]}
\displaystyle \text{Answer:}
\displaystyle \cos30^\circ=\frac{\sqrt3}{2},\quad\sin45^\circ=\frac{1}{\sqrt2},\quad\sin60^\circ=\frac{\sqrt3}{2}
\displaystyle \sec45^\circ=\sqrt2,\qquad\cot30^\circ=\sqrt3
\displaystyle \therefore \frac{2}{3}(\cos^4 30^\circ-\sin^4 45^\circ)-3(\sin^2 60^\circ-\sec^2 45^\circ)+\frac14\cot^2 30^\circ
\displaystyle =\frac23\left[\left(\frac{\sqrt3}{2}\right)^4-\left(\frac{1}{\sqrt2}\right)^4\right]-3\left[\left(\frac{\sqrt3}{2}\right)^2-(\sqrt2)^2\right]+\frac14(\sqrt3)^2
\displaystyle =\frac23\left(\frac{9}{16}-\frac14\right)-3\left(\frac34-2\right)+\frac34
\displaystyle =\frac23\left(\frac5{16}\right)-3\left(-\frac54\right)+\frac34
\displaystyle =\frac5{24}+\frac{15}{4}+\frac34
\displaystyle =\frac5{24}+\frac{90}{24}+\frac{18}{24}=\frac{113}{24}
\displaystyle \therefore \text{The required value is }\frac{113}{24}.

\displaystyle \text{(ii) }\frac{3\cos55^\circ}{7\sin35^\circ}-\frac{4(\cos70^\circ\mathrm{cosec}\,20^\circ)}{7(\tan5^\circ\tan25^\circ\tan45^\circ\tan65^\circ\tan85^\circ)}   \hfill\text{[CBSE 2007]}
\displaystyle \text{Answer:}
\displaystyle \cos55^\circ=\sin35^\circ,\qquad\cos70^\circ=\sin20^\circ
\displaystyle \tan85^\circ=\cot5^\circ,\quad\tan65^\circ=\cot25^\circ,\quad\tan45^\circ=1
\displaystyle \therefore \cos70^\circ\mathrm{cosec}\,20^\circ=\sin20^\circ\times\frac{1}{\sin20^\circ}=1
\displaystyle \text{and }\tan5^\circ\tan25^\circ\tan45^\circ\tan65^\circ\tan85^\circ
\displaystyle =(\tan5^\circ\cot5^\circ)(\tan25^\circ\cot25^\circ)(1)=1
\displaystyle \therefore \frac{3\cos55^\circ}{7\sin35^\circ}-\frac{4(\cos70^\circ\mathrm{cosec}\,20^\circ)}{7(\tan5^\circ\tan25^\circ\tan45^\circ\tan65^\circ\tan85^\circ)}
\displaystyle =\frac37-\frac47=-\frac17
\displaystyle \therefore \text{The required value is }-\frac17.

\displaystyle \text{(iii) }\frac{\cos58^\circ}{\sin32^\circ}+\frac{\sin22^\circ}{\cos68^\circ}-\frac{\cos38^\circ\mathrm{cosec}\,52^\circ}{\tan18^\circ\tan35^\circ\tan60^\circ\tan72^\circ\tan55^\circ}   \hfill\text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \cos58^\circ=\sin32^\circ,\qquad\sin22^\circ=\cos68^\circ
\displaystyle \sin52^\circ=\cos38^\circ\quad\therefore\quad\cos38^\circ\mathrm{cosec}\,52^\circ=1
\displaystyle \tan72^\circ=\cot18^\circ,\qquad\tan55^\circ=\cot35^\circ
\displaystyle \therefore \tan18^\circ\tan35^\circ\tan60^\circ\tan72^\circ\tan55^\circ
\displaystyle =(\tan18^\circ\cot18^\circ)(\tan35^\circ\cot35^\circ)\tan60^\circ=\sqrt3
\displaystyle \therefore \frac{\cos58^\circ}{\sin32^\circ}+\frac{\sin22^\circ}{\cos68^\circ}-\frac{\cos38^\circ\mathrm{cosec}\,52^\circ}{\tan18^\circ\tan35^\circ\tan60^\circ\tan72^\circ\tan55^\circ}
\displaystyle =1+1-\frac{1}{\sqrt3}
\displaystyle =2-\frac{\sqrt3}{3}=\frac{6-\sqrt3}{3}
\displaystyle \therefore \text{The required value is }\frac{6-\sqrt3}{3}.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If }2\theta+45^\circ\text{ and }30^\circ-\theta\text{ are acute angles, find the degree measure}
\displaystyle \text{of }\theta\text{ satisfying }\sin(2\theta+45^\circ)=\cos(30^\circ-\theta).
\displaystyle \text{Answer:}
\displaystyle \sin(2\theta+45^\circ)=\cos(30^\circ-\theta)
\displaystyle \text{Since }2\theta+45^\circ\text{ and }30^\circ-\theta\text{ are acute angles, they are complementary.}
\displaystyle \therefore (2\theta+45^\circ)+(30^\circ-\theta)=90^\circ
\displaystyle \theta+75^\circ=90^\circ
\displaystyle \therefore \theta=15^\circ.
\displaystyle \text{Also, }2\theta+45^\circ=75^\circ\text{ and }30^\circ-\theta=15^\circ,\text{ which are acute angles.}
\displaystyle \therefore \text{The required degree measure of }\theta\text{ is }15^\circ.
\displaystyle \\

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