\displaystyle \text{Note: Trigonometrical Ratios of Complementary Angles}
\displaystyle \text{For }\theta<90^{\circ},
\displaystyle \text{(i) }\sin(90^{\circ}-\theta)=\cos\theta \hspace{1.0cm} \text{(ii) }\cos(90^{\circ}-\theta)=\sin\theta
\displaystyle \text{(iii) }\tan(90^{\circ}-\theta)=\cot\theta \hspace{1.0cm} \text{(iv) }\cot(90^{\circ}-\theta)=\tan\theta
\displaystyle \text{(v) }\sec(90^{\circ}-\theta)=\mathrm{cosec}\,\theta \hspace{1.0cm} \text{(vi) }\mathrm{cosec}(90^{\circ}-\theta)=\sec\theta
\displaystyle \\

Evaluate:

\displaystyle \text{Question 1: } \frac{\cos 22^{\circ}}{\sin 68^{\circ}}
\displaystyle \text{Answer:}
\displaystyle \frac{\cos 22^{\circ}}{\sin 68^{\circ}}=\frac{\cos(90^{\circ}-68^{\circ})}{\sin 68^{\circ}}=\frac{\sin 68^{\circ}}{\sin 68^{\circ}}=1
\displaystyle \\

\displaystyle \text{Question 2: } \frac{\tan 47^{\circ}}{\cot 43^{\circ}}
\displaystyle \text{Answer:}
\displaystyle \frac{\tan 47^{\circ}}{\cot 43^{\circ}}=\frac{\tan(90^{\circ}-43^{\circ})}{\cot 43^{\circ}}=\frac{\cot 43^{\circ}}{\cot 43^{\circ}}=1
\displaystyle \\

\displaystyle \text{Question 3: } \frac{\sec 75^{\circ}}{\mathrm{cosec}\,15^{\circ}}
\displaystyle \text{Answer:}
\displaystyle \frac{\sec 75^{\circ}}{\mathrm{cosec}\,15^{\circ}}=\frac{\sec(90^{\circ}-15^{\circ})}{\mathrm{cosec}\,15^{\circ}}=\frac{\mathrm{cosec}\,15^{\circ}}{\mathrm{cosec}\,15^{\circ}}=1
\displaystyle \\

\displaystyle \text{Question 4: } \frac{\cos 55^{\circ}}{\sin 35^{\circ}}+\frac{\cot 35^{\circ}}{\tan 55^{\circ}}
\displaystyle \text{Answer:}
\displaystyle \frac{\cos 55^{\circ}}{\sin 35^{\circ}}+\frac{\cot 35^{\circ}}{\tan 55^{\circ}}=\frac{\cos(90^{\circ}-35^{\circ})}{\sin 35^{\circ}}+\frac{\cot(90^{\circ}-35^{\circ})}{\tan 55^{\circ}}=\frac{\sin 35^{\circ}}{\sin 35^{\circ}}+\frac{\tan 55^{\circ}}{\tan 55^{\circ}}=2
\displaystyle \\

\displaystyle \text{Question 5: } \cos^2 40^{\circ}+\cos^2 50^{\circ}
\displaystyle \text{Answer:}
\displaystyle \cos^2 40^{\circ}+\cos^2 50^{\circ}=(\cos(90^{\circ}-50^{\circ}))^2+\cos^2 50^{\circ}=\sin^2 50^{\circ}+\cos^2 50^{\circ}=1
\displaystyle \\

\displaystyle \text{Question 6: } \sec^2 18^{\circ}-\cot^2 72^{\circ}
\displaystyle \text{Answer:}
\displaystyle \sec^2 18^{\circ}-\cot^2 72^{\circ}=(\sec(90^{\circ}-72^{\circ}))^2-\cot^2 72^{\circ}=\mathrm{cosec}^2 72^{\circ}-\cot^2 72^{\circ}=1
\displaystyle \\

\displaystyle \text{Question 7: } \sin 15^{\circ}\cdot\cos 75^{\circ}+\cos 15^{\circ}\cdot\sin 75^{\circ}
\displaystyle \text{Answer:}
\displaystyle \sin 15^{\circ}\cdot\cos 75^{\circ}+\cos 15^{\circ}\cdot\sin 75^{\circ}
\displaystyle =\sin(90^{\circ}-75^{\circ})\cdot\cos 75^{\circ}+\cos(90^{\circ}-75^{\circ})\cdot\sin 75^{\circ}
\displaystyle =\cos^2 75^{\circ}+\sin^2 75^{\circ}=1
\displaystyle \\

\displaystyle \text{Question 8: } \sin 42^{\circ}\cdot\sin 48^{\circ}-\cos 42^{\circ}\cdot\cos 48^{\circ}
\displaystyle \text{Answer:}
\displaystyle \sin 42^{\circ}\cdot\sin 48^{\circ}-\cos 42^{\circ}\cdot\cos 48^{\circ}
\displaystyle =\sin 42^{\circ}\cdot\sin(90^{\circ}-42^{\circ})-\cos 42^{\circ}\cdot\cos(90^{\circ}-42^{\circ})
\displaystyle =\sin 42^{\circ}\cdot\cos 42^{\circ}-\cos 42^{\circ}\cdot\sin 42^{\circ}=0
\displaystyle \\

\displaystyle \text{Question 9: } \sin(90^{\circ}-A)\cos A+\cos(90^{\circ}-A)\sin A
\displaystyle \text{Answer:}
\displaystyle \sin(90^{\circ}-A)\cos A+\cos(90^{\circ}-A)\sin A=\cos^2 A+\sin^2 A=1
\displaystyle \\

\displaystyle \text{Question 10: } \sin^2 35^{\circ}+\sin^2 55^{\circ}
\displaystyle \text{Answer:}
\displaystyle \sin^2 35^{\circ}+\sin^2 55^{\circ}=\{\sin(90^{\circ}-55^{\circ})\}^2+\sin^2 55^{\circ}=\cos^2 55^{\circ}+\sin^2 55^{\circ}=1
\displaystyle \\

\displaystyle \text{Question 11: } \frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}-2
\displaystyle \text{Answer:}
\displaystyle \frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}-2=\frac{\cot(90^{\circ}-36^{\circ})}{\tan 36^{\circ}}+\frac{\tan(90^{\circ}-70^{\circ})}{\cot 70^{\circ}}-2
\displaystyle =\frac{\tan 36^{\circ}}{\tan 36^{\circ}}+\frac{\cot 70^{\circ}}{\cot 70^{\circ}}-2=2-2=0
\displaystyle \\

\displaystyle \text{Question 12: } \frac{2\tan 53^{\circ}}{\cot 37^{\circ}}-\frac{\cot 80^{\circ}}{\tan 10^{\circ}} \hspace{1.0cm} [\text{ICSE }2006]
\displaystyle \text{Answer:}
\displaystyle \frac{2\tan 53^{\circ}}{\cot 37^{\circ}}-\frac{\cot 80^{\circ}}{\tan 10^{\circ}}=\frac{2\tan(90^{\circ}-37^{\circ})}{\cot 37^{\circ}}-\frac{\cot(90^{\circ}-10^{\circ})}{\tan 10^{\circ}}
\displaystyle =\frac{2\cot 37^{\circ}}{\cot 37^{\circ}}-\frac{\tan 10^{\circ}}{\tan 10^{\circ}}=2-1=1
\displaystyle \\

\displaystyle \text{Question 13: } \cos^2 25^{\circ}+\cos^2 65^{\circ}-\tan^2 45^{\circ}
\displaystyle \text{Answer:}
\displaystyle \cos^2 25^{\circ}+\cos^2 65^{\circ}-\tan^2 45^{\circ}
\displaystyle =\cos^2(90^{\circ}-65^{\circ})+\cos^2 65^{\circ}-\tan^2 45^{\circ}
\displaystyle =\sin^2 65^{\circ}+\cos^2 65^{\circ}-1=1-1=0
\displaystyle \\

\displaystyle \text{Question 14: } \frac{\cos^2 32^{\circ}+\cos^2 58^{\circ}}{\sin^2 59^{\circ}+\sin^2 31^{\circ}}
\displaystyle \text{Answer:}
\displaystyle \frac{\cos^2 32^{\circ}+\cos^2 58^{\circ}}{\sin^2 59^{\circ}+\sin^2 31^{\circ}}
\displaystyle =\frac{\cos^2(90^{\circ}-58^{\circ})+\cos^2 58^{\circ}}{\sin^2(90^{\circ}-31^{\circ})+\sin^2 31^{\circ}}
\displaystyle =\frac{\sin^2 58^{\circ}+\cos^2 58^{\circ}}{\cos^2 31^{\circ}+\sin^2 31^{\circ}}=\frac{1}{1}=1
\displaystyle \\

\displaystyle \text{Question 15: } \left(\frac{\sin 77^{\circ}}{\cos 13^{\circ}}\right)^2+\left(\frac{\cos 77^{\circ}}{\sin 13^{\circ}}\right)^2-2\cos^2 45^{\circ}
\displaystyle \text{Answer:}
\displaystyle \left(\frac{\sin 77^{\circ}}{\cos 13^{\circ}}\right)^2+\left(\frac{\cos 77^{\circ}}{\sin 13^{\circ}}\right)^2-2\cos^2 45^{\circ}
\displaystyle =\left(\frac{\sin(90^{\circ}-13^{\circ})}{\cos 13^{\circ}}\right)^2+\left(\frac{\cos(90^{\circ}-13^{\circ})}{\sin 13^{\circ}}\right)^2-2\cos^2 45^{\circ}
\displaystyle =\left(\frac{\cos 13^{\circ}}{\cos 13^{\circ}}\right)^2+\left(\frac{\sin 13^{\circ}}{\sin 13^{\circ}}\right)^2-2\left(\frac{1}{\sqrt{2}}\right)^2
\displaystyle =1+1-2\left(\frac{1}{2}\right)=1
\displaystyle \\

\displaystyle \text{Question 16: } \cos^2 26^{\circ}+\cos 64^{\circ}\cdot\sin 26^{\circ}+\frac{\tan 36^{\circ}}{\cot 54^{\circ}} \hspace{1.0cm} [\text{ICSE }2012]
\displaystyle \text{Answer:}
\displaystyle \cos^2 26^{\circ}+\cos 64^{\circ}\cdot\sin 26^{\circ}+\frac{\tan 36^{\circ}}{\cot 54^{\circ}}
\displaystyle =\cos^2 26^{\circ}+\cos(90^{\circ}-26^{\circ})\cdot\sin 26^{\circ}+\frac{\tan(90^{\circ}-54^{\circ})}{\cot 54^{\circ}}
\displaystyle =\cos^2 26^{\circ}+\sin^2 26^{\circ}+\frac{\cot 54^{\circ}}{\cot 54^{\circ}}
\displaystyle =1+1=2
\displaystyle \\

\displaystyle \text{Question 17: } 3\cdot\frac{\sin 72^{\circ}}{\cos 18^{\circ}}-\frac{\sec 32^{\circ}}{\mathrm{cosec}\,58^{\circ}}
\displaystyle \text{Answer:}
\displaystyle 3\cdot\frac{\sin 72^{\circ}}{\cos 18^{\circ}}-\frac{\sec 32^{\circ}}{\mathrm{cosec}\,58^{\circ}}
\displaystyle =3\cdot\frac{\sin(90^{\circ}-18^{\circ})}{\cos 18^{\circ}}-\frac{\mathrm{cosec}(90^{\circ}-32^{\circ})}{\mathrm{cosec}\,58^{\circ}}
\displaystyle =3\cdot\frac{\cos 18^{\circ}}{\cos 18^{\circ}}-\frac{\mathrm{cosec}\,58^{\circ}}{\mathrm{cosec}\,58^{\circ}}
\displaystyle =3-1=2
\displaystyle \\

\displaystyle \text{Question 18: } 3\cos 80^{\circ}\cdot\mathrm{cosec}\,10^{\circ}+2\sin 59^{\circ}\cdot\sec 31^{\circ} \hspace{1.0cm} [\text{ICSE }2013]
\displaystyle \text{Answer:}
\displaystyle 3\cos 80^{\circ}\cdot\mathrm{cosec}\,10^{\circ}+2\sin 59^{\circ}\cdot\sec 31^{\circ}
\displaystyle =3\cos 80^{\circ}\cdot\mathrm{cosec}(90^{\circ}-80^{\circ})+2\sin 59^{\circ}\cdot\sec(90^{\circ}-59^{\circ})
\displaystyle =3\cos 80^{\circ}\cdot\sec 80^{\circ}+2\sin 59^{\circ}\cdot\mathrm{cosec}\,59^{\circ}
\displaystyle =3+2=5
\displaystyle \\

\displaystyle \text{Question 19: } \frac{\sin 80^{\circ}}{\cos 10^{\circ}}+\sin 59^{\circ}\cdot\sec 31^{\circ} \hspace{1.0cm} [\text{ICSE }2007]
\displaystyle \text{Answer:}
\displaystyle \frac{\sin 80^{\circ}}{\cos 10^{\circ}}+\sin 59^{\circ}\cdot\sec 31^{\circ}
\displaystyle =\frac{\sin(90^{\circ}-10^{\circ})}{\cos 10^{\circ}}+\sin 59^{\circ}\cdot\sec(90^{\circ}-59^{\circ})
\displaystyle =\frac{\cos 10^{\circ}}{\cos 10^{\circ}}+\sin 59^{\circ}\cdot\mathrm{cosec}\,59^{\circ}
\displaystyle =1+1=2
\displaystyle \\

\displaystyle \text{Question 20: } \tan(55^{\circ}-A)-\cot(35^{\circ}+A)
\displaystyle \text{Answer:}
\displaystyle \tan(55^{\circ}-A)-\cot(35^{\circ}+A)
\displaystyle =\tan\left(90^{\circ}-(35^{\circ}+A)\right)-\cot(35^{\circ}+A)
\displaystyle =\cot(35^{\circ}+A)-\cot(35^{\circ}+A)=0
\displaystyle \\

\displaystyle \text{Question 21: } \mathrm{cosec}(65^{\circ}+A)-\sec(25^{\circ}-A)
\displaystyle \text{Answer:}
\displaystyle \mathrm{cosec}(65^{\circ}+A)-\sec(25^{\circ}-A)
\displaystyle =\mathrm{cosec}\left(90^{\circ}-(25^{\circ}-A)\right)-\sec(25^{\circ}-A)
\displaystyle =\sec(25^{\circ}-A)-\sec(25^{\circ}-A)=0
\displaystyle \\

\displaystyle \text{Question 22: } 2\cdot\frac{\tan 57^{\circ}}{\cot 33^{\circ}}-\frac{\cot 70^{\circ}}{\tan 20^{\circ}}-\sqrt{2}\cos 45^{\circ}
\displaystyle \text{Answer:}
\displaystyle 2\cdot\frac{\tan 57^{\circ}}{\cot 33^{\circ}}-\frac{\cot 70^{\circ}}{\tan 20^{\circ}}-\sqrt{2}\cos 45^{\circ}
\displaystyle =2\cdot\frac{\tan(90^{\circ}-33^{\circ})}{\cot 33^{\circ}}-\frac{\cot(90^{\circ}-20^{\circ})}{\tan 20^{\circ}}-\sqrt{2}\cos 45^{\circ}
\displaystyle =2\cdot\frac{\cot 33^{\circ}}{\cot 33^{\circ}}-\frac{\tan 20^{\circ}}{\tan 20^{\circ}}-\sqrt{2}\cos 45^{\circ}
\displaystyle =2-1-\sqrt{2}\times\frac{1}{\sqrt{2}}
\displaystyle =2-1-1=0
\displaystyle \\

\displaystyle \text{Question 23: } \frac{\cot^2 41^{\circ}}{\tan^2 49^{\circ}}-2\frac{\sin^2 75^{\circ}}{\cos^2 15^{\circ}}
\displaystyle \text{Answer:}
\displaystyle \frac{\cot^2 41^{\circ}}{\tan^2 49^{\circ}}-2\frac{\sin^2 75^{\circ}}{\cos^2 15^{\circ}}
\displaystyle =\frac{\cot^2(90^{\circ}-49^{\circ})}{\tan^2 49^{\circ}}-2\frac{\sin^2(90^{\circ}-15^{\circ})}{\cos^2 15^{\circ}}
\displaystyle =\frac{\tan^2 49^{\circ}}{\tan^2 49^{\circ}}-2\frac{\cos^2 15^{\circ}}{\cos^2 15^{\circ}}
\displaystyle =1-2=-1
\displaystyle \\

\displaystyle \text{Question 24: } \frac{\cos 70^{\circ}}{\sin 20^{\circ}}+\frac{\cos 59^{\circ}}{\sin 31^{\circ}}-8\sin^2 30^{\circ}
\displaystyle \text{Answer:}
\displaystyle \frac{\cos 70^{\circ}}{\sin 20^{\circ}}+\frac{\cos 59^{\circ}}{\sin 31^{\circ}}-8\sin^2 30^{\circ}
\displaystyle =\frac{\cos(90^{\circ}-20^{\circ})}{\sin 20^{\circ}}+\frac{\cos(90^{\circ}-31^{\circ})}{\sin 31^{\circ}}-8\sin^2 30^{\circ}
\displaystyle =\frac{\sin 20^{\circ}}{\sin 20^{\circ}}+\frac{\sin 31^{\circ}}{\sin 31^{\circ}}-8\times\frac{1}{4}
\displaystyle =1+1-2=0
\displaystyle \\

\displaystyle \text{Question 25: } 14\sin 30^{\circ}+6\cos 60^{\circ}-5\tan 45^{\circ} \hspace{1.0cm} [\text{ICSE }2004]
\displaystyle \text{Answer:}
\displaystyle 14\sin 30^{\circ}+6\cos 60^{\circ}-5\tan 45^{\circ}
\displaystyle =14\sin(90^{\circ}-60^{\circ})+6\cos 60^{\circ}-5\tan 45^{\circ}
\displaystyle =14\cos 60^{\circ}+6\cos 60^{\circ}-5\tan 45^{\circ}
\displaystyle =20\cos 60^{\circ}-5\tan 45^{\circ}
\displaystyle =20\times\frac{1}{2}-5\times1
\displaystyle =10-5=5
\displaystyle \\

\displaystyle \text{Question 26: Show that } \tan 10^{\circ}\tan 15^{\circ}\tan 75^{\circ}\tan 80^{\circ}=1
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\tan 10^{\circ}\tan 15^{\circ}\tan 75^{\circ}\tan 80^{\circ}
\displaystyle =\tan(90^{\circ}-80^{\circ})\tan(90^{\circ}-75^{\circ})\tan 75^{\circ}\tan 80^{\circ}
\displaystyle =\cot 80^{\circ}\cot 75^{\circ}\tan 75^{\circ}\tan 80^{\circ}
\displaystyle =1=\text{RHS}.
\displaystyle \therefore \text{Hence proved.}
\displaystyle \\

\displaystyle \text{Question 27: Show that } \sin 42^{\circ}\cdot\sec 48^{\circ}+\cos 42^{\circ}\cdot\mathrm{cosec}\,48^{\circ}=2
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sin 42^{\circ}\cdot\sec 48^{\circ}+\cos 42^{\circ}\cdot\mathrm{cosec}\,48^{\circ}
\displaystyle =\sin 42^{\circ}\cdot\sec(90^{\circ}-42^{\circ})+\cos 42^{\circ}\cdot\mathrm{cosec}(90^{\circ}-42^{\circ})
\displaystyle =\sin 42^{\circ}\cdot\mathrm{cosec}\,42^{\circ}+\cos 42^{\circ}\cdot\sec 42^{\circ}
\displaystyle =1+1=2=\text{RHS}.
\displaystyle \therefore \text{Hence proved.}
\displaystyle \\

\displaystyle \text{Question 28: Show that } \frac{\sin 26^{\circ}}{\sec 64^{\circ}}+\frac{\cos 26^{\circ}}{\mathrm{cosec}\,64^{\circ}}=1
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\sin 26^{\circ}}{\sec 64^{\circ}}+\frac{\cos 26^{\circ}}{\mathrm{cosec}\,64^{\circ}}
\displaystyle =\frac{\sin 26^{\circ}}{\sec(90^{\circ}-26^{\circ})}+\frac{\cos 26^{\circ}}{\mathrm{cosec}(90^{\circ}-26^{\circ})}
\displaystyle =\frac{\sin 26^{\circ}}{\mathrm{cosec}\,26^{\circ}}+\frac{\cos 26^{\circ}}{\sec 26^{\circ}}
\displaystyle =\sin^2 26^{\circ}+\cos^2 26^{\circ}=1=\text{RHS}.
\displaystyle \therefore \text{Hence proved.}
\displaystyle \\

\displaystyle \text{Question 29: Express in terms of angles between }0^{\circ}\text{ and }45^{\circ}\text{: }\sin 59^{\circ}+\tan 63^{\circ}
\displaystyle \text{Answer:}
\displaystyle \sin 59^{\circ}+\tan 63^{\circ}=\sin(90^{\circ}-31^{\circ})+\tan(90^{\circ}-27^{\circ})=\cos 31^{\circ}+\cot 27^{\circ}
\displaystyle \\

\displaystyle \text{Question 30: Express in terms of angles between }0^{\circ}\text{ and }45^{\circ}\text{: }\mathrm{cosec}\,68^{\circ}+\cot 72^{\circ}
\displaystyle \text{Answer:}
\displaystyle \mathrm{cosec}\,68^{\circ}+\cot 72^{\circ}=\mathrm{cosec}(90^{\circ}-32^{\circ})+\cot(90^{\circ}-28^{\circ})=\sec 32^{\circ}+\tan 28^{\circ}
\displaystyle \\

\displaystyle \text{Question 31: Express in terms of angles between }0^{\circ}\text{ and }45^{\circ}\text{: }\cos 74^{\circ}+\sec 67^{\circ}
\displaystyle \text{Answer:}
\displaystyle \cos 74^{\circ}+\sec 67^{\circ}=\cos(90^{\circ}-16^{\circ})+\sec(90^{\circ}-23^{\circ})=\sin 16^{\circ}+\mathrm{cosec}\,23^{\circ}
\displaystyle \\

\displaystyle \text{Question 32: } \frac{\sin A}{\sin(90^{\circ}-A)}+\frac{\cos A}{\cos(90^{\circ}-A)}=\sec A\cdot\mathrm{cosec}\,A
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\sin A}{\sin(90^{\circ}-A)}+\frac{\cos A}{\cos(90^{\circ}-A)}
\displaystyle =\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}
\displaystyle =\frac{\sin^2 A+\cos^2 A}{\sin A\cos A}
\displaystyle =\frac{1}{\sin A\cos A}
\displaystyle =\sec A\cdot\mathrm{cosec}\,A=\text{RHS}.
\displaystyle \therefore \text{Hence proved.}
\displaystyle \\

\displaystyle \text{Question 33: } \sin A\cos A-\frac{\sin A\cos(90^{\circ}-A)\cos A}{\sec(90^{\circ}-A)}-\frac{\cos A\sin(90^{\circ}-A)\sin A}{\mathrm{cosec}(90^{\circ}-A)}=0
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sin A\cos A-\frac{\sin A\cos(90^{\circ}-A)\cos A}{\sec(90^{\circ}-A)}-\frac{\cos A\sin(90^{\circ}-A)\sin A}{\mathrm{cosec}(90^{\circ}-A)}
\displaystyle =\sin A\cos A-\frac{\sin^2 A\cos A}{\mathrm{cosec}\,A}-\frac{\cos^2 A\sin A}{\sec A}
\displaystyle =\sin A\cos A-\sin^3 A\cos A-\cos^3 A\sin A
\displaystyle =\sin A\cos A-\sin A\cos A(\sin^2 A+\cos^2 A)
\displaystyle =\sin A\cos A-\sin A\cos A
\displaystyle =0=\text{RHS}.
\displaystyle \therefore \text{Hence proved.}
\displaystyle \\

\displaystyle \text{Question 34: For }\triangle ABC,\text{ show that }\sin\left(\frac{A+B}{2}\right)=\cos\left(\frac{C}{2}\right)
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sin\left(\frac{A+B}{2}\right)
\displaystyle =\sin\left(\frac{180^{\circ}-C}{2}\right)
\displaystyle =\sin\left(90^{\circ}-\frac{C}{2}\right)
\displaystyle =\cos\frac{C}{2}=\text{RHS}.
\displaystyle \therefore \text{Hence proved.}
\displaystyle \\

\displaystyle \text{Question 35: For }\triangle ABC,\text{ show that }\tan\left(\frac{B+C}{2}\right)=\cot\left(\frac{A}{2}\right)
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\tan\left(\frac{B+C}{2}\right)
\displaystyle =\tan\left(\frac{180^{\circ}-A}{2}\right)
\displaystyle =\tan\left(90^{\circ}-\frac{A}{2}\right)
\displaystyle =\cot\frac{A}{2}=\text{RHS}.
\displaystyle \therefore \text{Hence proved.}
\displaystyle \\

\displaystyle \text{Question 36: In }\triangle ABC,\;\angle B=90^{\circ}.\text{ Find the value of }\frac{\sec A\cdot\mathrm{cosec}\,A-\tan A\cdot\cot C}{\sin B}
\displaystyle \text{Answer:}
\displaystyle \frac{\sec A\cdot\mathrm{cosec}\,A-\tan A\cdot\cot C}{\sin B}
\displaystyle =\frac{\sec A\cdot\mathrm{cosec}(90^{\circ}-A)-\tan A\cdot\cot(90^{\circ}-A)}{\sin 90^{\circ}}
\displaystyle =\frac{\sec A\cdot\sec A-\tan A\cdot\tan A}{1}
\displaystyle =\sec^2 A-\tan^2 A
\displaystyle =1
\displaystyle \\

\displaystyle \text{Question 37: Find }x\text{ if }\sin x=\sin 60^{\circ}\cos 30^{\circ}-\cos 60^{\circ}\sin 30^{\circ}
\displaystyle \text{Answer:}
\displaystyle \sin x=\sin 60^{\circ}\cos 30^{\circ}-\cos 60^{\circ}\sin 30^{\circ}
\displaystyle =\sin(60^{\circ}-30^{\circ})
\displaystyle =\sin 30^{\circ}
\displaystyle \Rightarrow x=30^{\circ}
\displaystyle \\

\displaystyle \text{Question 38: Find }x\text{ if }\sin x=\sin 60^{\circ}\cos 30^{\circ}+\cos 60^{\circ}\sin 30^{\circ}
\displaystyle \text{Answer:}
\displaystyle \sin x=\sin 60^{\circ}\cos 30^{\circ}+\cos 60^{\circ}\sin 30^{\circ}
\displaystyle \Rightarrow \sin x=\sin 60^{\circ}\cos(90^{\circ}-60^{\circ})+\cos 60^{\circ}\sin(90^{\circ}-60^{\circ})
\displaystyle \Rightarrow \sin x=\sin^2 60^{\circ}+\cos^2 60^{\circ}
\displaystyle \Rightarrow \sin x=1
\displaystyle \Rightarrow x=90^{\circ}
\displaystyle \\

\displaystyle \text{Question 39: Find }x\text{ if }\cos x=\cos 60^{\circ}\cos 30^{\circ}-\sin 60^{\circ}\sin 30^{\circ}
\displaystyle \text{Answer:}
\displaystyle \cos x=\cos 60^{\circ}\cos 30^{\circ}-\sin 60^{\circ}\sin 30^{\circ}
\displaystyle \Rightarrow \cos x=\cos 60^{\circ}\cos(90^{\circ}-60^{\circ})-\sin 60^{\circ}\sin(90^{\circ}-60^{\circ})
\displaystyle \Rightarrow \cos x=\cos 60^{\circ}\sin 60^{\circ}-\sin 60^{\circ}\cos 60^{\circ}
\displaystyle \Rightarrow \cos x=0
\displaystyle \Rightarrow x=90^{\circ}
\displaystyle \\

\displaystyle \text{Question 40: Find the value of }x\text{ if }\tan x=\frac{\tan 60^{\circ}-\tan 30^{\circ}}{1+\tan 60^{\circ}\tan 30^{\circ}}
\displaystyle \text{Answer:}
\displaystyle \tan x=\frac{\tan 60^{\circ}-\tan 30^{\circ}}{1+\tan 60^{\circ}\tan 30^{\circ}}
\displaystyle =\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3}\times\frac{1}{\sqrt{3}}}
\displaystyle =\frac{\frac{3-1}{\sqrt{3}}}{2}
\displaystyle =\frac{1}{\sqrt{3}}
\displaystyle =\tan 30^{\circ}
\displaystyle \Rightarrow x=30^{\circ}
\displaystyle \\

\displaystyle \text{Question 41: Find the value of }x\text{ if }\sin 2x=2\sin 45^{\circ}\cos 45^{\circ}
\displaystyle \text{Answer:}
\displaystyle \sin 2x=2\sin 45^{\circ}\cos 45^{\circ}
\displaystyle \Rightarrow \sin 2x=2\times\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}}
\displaystyle \Rightarrow \sin 2x=1
\displaystyle \Rightarrow 2x=90^{\circ}
\displaystyle \Rightarrow x=45^{\circ}
\displaystyle \\

\displaystyle \text{Question 42: Find the value of }x\text{ if }\sin 3x=2\sin 30^{\circ}\cos 30^{\circ}
\displaystyle \text{Answer:}
\displaystyle \sin 3x=2\sin 30^{\circ}\cos 30^{\circ}
\displaystyle \Rightarrow \sin 3x=2\times\frac{1}{2}\times\frac{\sqrt{3}}{2}
\displaystyle \Rightarrow \sin 3x=\frac{\sqrt{3}}{2}
\displaystyle \Rightarrow 3x=60^{\circ}
\displaystyle \Rightarrow x=20^{\circ}
\displaystyle \\

\displaystyle \text{Question 43: Find }x\text{ if }\cos(2x-6^{\circ})=\cos^2 30^{\circ}-\cos^2 60^{\circ}
\displaystyle \text{Answer:}
\displaystyle \cos(2x-6^{\circ})=\cos^2 30^{\circ}-\cos^2 60^{\circ}
\displaystyle \Rightarrow \cos(2x-6^{\circ})=\sin^2 60^{\circ}-\cos^2 60^{\circ}
\displaystyle \Rightarrow \cos(2x-6^{\circ})=\left(\frac{\sqrt{3}}{2}\right)^2-\left(\frac{1}{2}\right)^2
\displaystyle \Rightarrow \cos(2x-6^{\circ})=\frac{1}{2}
\displaystyle \Rightarrow 2x-6^{\circ}=60^{\circ}
\displaystyle \Rightarrow x=33^{\circ}
\displaystyle \\

\displaystyle \text{Question 44: Find the value of }A\text{ where }0^{\circ}\leq A\leq90^{\circ}\text{: }\sin(90^{\circ}-3A)\cdot\mathrm{cosec}\,42^{\circ}=1
\displaystyle \text{Answer:}
\displaystyle \sin(90^{\circ}-3A)\cdot\mathrm{cosec}\,42^{\circ}=1
\displaystyle \Rightarrow \sin(90^{\circ}-3A)=\sin42^{\circ}
\displaystyle \Rightarrow \cos3A=\sin42^{\circ}=\cos48^{\circ}
\displaystyle \Rightarrow 3A=48^{\circ}
\displaystyle \Rightarrow A=16^{\circ}
\displaystyle \\

\displaystyle \text{Question 45: Find the value of }A\text{ where }0^{\circ}\leq A\leq90^{\circ}\text{: }\cos(90^{\circ}-A)\cdot\sec77^{\circ}=1
\displaystyle \text{Answer:}
\displaystyle \cos(90^{\circ}-A)\cdot\sec77^{\circ}=1
\displaystyle \Rightarrow \cos(90^{\circ}-A)=\cos77^{\circ}
\displaystyle \Rightarrow \sin A=\cos77^{\circ}=\sin13^{\circ}
\displaystyle \Rightarrow A=13^{\circ}
\displaystyle \\

\displaystyle \text{Question 46: Prove that }\frac{\cos(90^{\circ}-A)\cdot\cos A}{\cot A}=1-\cos^2 A
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\cos(90^{\circ}-A)\cdot\cos A}{\cot A}
\displaystyle =\frac{\sin A\cos A}{\frac{\cos A}{\sin A}}
\displaystyle =\frac{\sin A\cos A\cdot\sin A}{\cos A}
\displaystyle =\sin^2 A
\displaystyle =1-\cos^2 A=\text{RHS}.
\displaystyle \therefore \text{Hence proved.}
\displaystyle \\

\displaystyle \text{Question 47: Prove that }\frac{\sin A\cdot\sin(90^{\circ}-A)}{\tan(90^{\circ}-A)}=1-\sin^2 A
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\sin A\cdot\sin(90^{\circ}-A)}{\tan(90^{\circ}-A)}
\displaystyle =\frac{\sin A\cos A}{\cot A}
\displaystyle =\frac{\sin A\cos A}{\frac{\cos A}{\sin A}}
\displaystyle =\frac{\sin A\cos A\cdot\sin A}{\cos A}
\displaystyle =\sin^2 A
\displaystyle =1-\cos^2 A

\displaystyle \text{Question 48: Evaluate }\frac{\sin 35^{\circ}\cdot\cos 55^{\circ}+\cos 35^{\circ}\cdot\sin 55^{\circ}}{\mathrm{cosec}^2 10^{\circ}-\tan^2 80^{\circ}} \hspace{1.0cm} [\text{ICSE }2010]
\displaystyle \text{Answer:}
\displaystyle \frac{\sin 35^{\circ}\cdot\cos 55^{\circ}+\cos 35^{\circ}\cdot\sin 55^{\circ}}{\mathrm{cosec}^2 10^{\circ}-\tan^2 80^{\circ}}
\displaystyle =\frac{\sin 35^{\circ}\cdot\cos(90^{\circ}-35^{\circ})+\cos 35^{\circ}\cdot\sin(90^{\circ}-35^{\circ})}{\mathrm{cosec}^2 10^{\circ}-\tan^2(90^{\circ}-10^{\circ})}
\displaystyle =\frac{\sin^2 35^{\circ}+\cos^2 35^{\circ}}{\mathrm{cosec}^2 10^{\circ}-\cot^2 10^{\circ}}
\displaystyle =\frac{1}{1}=1
\displaystyle \\

\displaystyle \text{Question 49: Without using trigonometric tables, evaluate }\sin^2 34^{\circ}+\sin^2 56^{\circ}+2\tan 18^{\circ}\cdot\tan 72^{\circ}-\cot^2 30^{\circ} \hspace{1.0cm} [\text{ICSE }2014]
\displaystyle \text{Answer:}
\displaystyle \sin^2 34^{\circ}+\sin^2 56^{\circ}+2\tan 18^{\circ}\cdot\tan 72^{\circ}-\cot^2 30^{\circ}
\displaystyle =\sin^2 34^{\circ}+\cos^2 34^{\circ}+2\tan 18^{\circ}\cdot\cot 18^{\circ}-\cot^2 30^{\circ}
\displaystyle =1+2-(\sqrt{3})^2
\displaystyle =3-3=0
\displaystyle \\

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