\displaystyle \text{Question 1: } \frac{\sec A -1 }{\sec A + 1} = \frac{1- \cos A }{1+ \cos A} \hspace{1.0cm} \text{[ICSE 2007]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS } =  \frac{\sec A -1 }{\sec A + 1}
\displaystyle =  \frac{\frac{1}{\cos A} -1 }{\frac{1}{\cos A} + 1}
\displaystyle =  \frac{1- \cos A }{1+ \cos A} = \text{ RHS. Hence Proved.}
\displaystyle \\
\displaystyle \text{Question 2: } (1-\tan A)^2 + (1 + \tan A)^2 = 2 \sec^2 A \hspace{1.0cm} \text{[ICSE 2005]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS } =  (1-\tan A)^2 + (1 + \tan A)^2
\displaystyle =  (1- \frac{\sin A}{\cos A})^2 + (1+\frac{\sin A}{\cos A})^2
\displaystyle =  \frac{(\cos A - \sin A)^2}{\cos^2 A} + \frac{(\cos A + \sin A)^2}{\cos^2 A}
\displaystyle =  \frac{\cos^2 A + \sin^2 A - 2 \cos A . \sin A+ \cos^2 A + \sin^2 A + 2 \cos A . \sin A}{\cos^2 A}
\displaystyle =  \frac{2}{\cos^2 A} = 2 \sec^2 A = \text{ RHS. Hence Proved.}
OR
\displaystyle \text{LHS } =  (1-\tan A)^2 + (1 + \tan A)^2
\displaystyle =  1 + \tan^2 A - 2 \tan A + 1 + \tan^2 A + 2 \tan A
\displaystyle  =  2 + 2\tan^2 A 
\displaystyle  =  2(1  + \tan^2 A) = 2 \sec^2 A = \text{ RHS. Hence Proved.}
\displaystyle \\
\displaystyle \text{Question 3: } \frac{\sin A}{1 + \cos A} = \mathrm{cosec} A - \cot A \hspace{1.0cm} \text{[ICSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{RHS } = \mathrm{cosec} A - \cot A
\displaystyle = \frac{1}{\sin A} - \frac{\cos A}{\sin A}
\displaystyle = \frac{1- \cos A}{\sin A}
\displaystyle = \frac{1- \cos A}{\sin A} \times \frac{1+ \cos A}{1+ \cos A}
\displaystyle = \frac{1 - \cos^2 A}{\sin A (1 + \cos A)}
\displaystyle = \frac{\sin A}{1 + \cos A} = \text{ LHS. Hence Proved.}
\displaystyle \\
\displaystyle \text{Question 4: } \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \mathrm{cosec} A - \cot A \hspace{1.0cm} \text{[ICSE 2000]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS } = \sqrt{\frac{1 - \cos A}{1 + \cos A}}
\displaystyle = \sqrt{\frac{1 - \cos A}{1 + \cos A} \times \frac{1 - \cos A}{1 - \cos A}}
\displaystyle = \sqrt{\frac{(1 - \cos A)^2}{1 - \cos^2 A}}
\displaystyle = \sqrt{\frac{(1 - \cos A)^2}{\sin^2 A}}
\displaystyle = \frac{1- \cos A}{\sin A}
\displaystyle = \mathrm{cosec} A - \cot A = \text{ RHS. Hence Proved.}
\displaystyle \\
\displaystyle \text{Question 5: } \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\sin A}{1 + \cos A} \hspace{1.0cm} \text{[ICSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS } = \sqrt{\frac{1 - \cos A}{1 + \cos A}}
\displaystyle = \sqrt{\frac{1 - \cos A}{1 + \cos A} \times \frac{1 + \cos A}{1 + \cos A}}
\displaystyle = \sqrt{\frac{1 - \cos^2 A}{(1 + \cos A)^2}}
\displaystyle = \frac{\sin A}{1 + \cos A} = \text{ RHS. Hence Proved.}
\displaystyle \\
\displaystyle \text{Question 6: } 1 - \frac{\cos^2 A}{1 + \sin A} = \sin A \hspace{1.0cm} \text{[ICSE 2001]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS } = \frac{\cos^2 A}{1 + \sin A}
\displaystyle = \frac{1 + \sin A - \cos^2 A}{1 + \sin A}
\displaystyle = \frac{\sin A + \sin^2 A}{1 + \sin A}
\displaystyle = \frac{\sin A(1 + \sin A)}{1 + \sin A}
\displaystyle = \sin A = \text{ RHS. Hence Proved.}
\displaystyle \\
\displaystyle \text{Question 7: } \frac{1}{\sin A + \cos A} + \frac{1}{\sin A - \cos A} = \frac{2 \sin A}{1 - 2 \cos^2 A} \hspace{1.0cm} \text{[ICSE 2002]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS } = \frac{1}{\sin A + \cos A} + \frac{1}{\sin A - \cos A}
\displaystyle = \frac{\sin A - \cos A + \sin A + \cos A}{\sin^2 A - \cos^2 A}
\displaystyle = \frac{\sin A - \cos A + \sin A + \cos A}{1 - \cos^2 A - \cos^2 A}
\displaystyle = \frac{2 \sin A}{1 - 2 \cos^2 A}  = \text{ RHS. Hence Proved.}
\displaystyle \\
\displaystyle \text{Question 8: } \frac{\sin \theta . \tan \theta}{1 - \cos \theta} = 1 + \sec \theta \hspace{1.0cm} \text{[ICSE 2006]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS } = \frac{\sin \theta . \tan \theta}{1 - \cos \theta}
\displaystyle = \frac{\sin^2 \theta}{\cos \theta (1 - \cos \theta)}
\displaystyle = \frac{(1-\cos \theta)(1 + \cos \theta)}{\cos \theta (1 - \cos \theta)}
\displaystyle = \frac{1 + \cos \theta}{\cos \theta}
\displaystyle = \sec \theta + 1 = \text{ RHS. Hence Proved.}
\displaystyle \\
\displaystyle \text{Question 9: } \frac{2 \tan 53^o}{\cot 37^o}-\frac{\cot 80^o}{\tan 10^o} \hspace{1.0cm} \text{[ICSE 2006]}
\displaystyle \text{Answer:}
\displaystyle \frac{2 \tan 53^o}{\cot 37^o}-\frac{\cot 80^o}{\tan 10^o}
\displaystyle = \frac{2 \tan (90^o - 37^o)}{\cot 37^o}-\frac{\cot (90^o - 10^o)}{\tan 10^o}
\displaystyle = \frac{2 \cot 37^o}{\cot 37^o}-\frac{\tan 10^o}{\tan 10^o} = 1
\displaystyle \\
\displaystyle \text{Question 10: } \cos^2 26^o + \cos 64^o.\sin 26^o + \frac{\tan 36^o}{\cot 54^o} \hspace{1.0cm} \text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \cos^2 26^o + \cos 64^o.\sin 26^o + \frac{\tan 36^o}{\cot 54^o}
\displaystyle = \cos^2 26^o + \cos (90^o - 26^o).\sin 26^o + \frac{\tan (90^o - 54^o)}{\cot 54^o}
\displaystyle = \cos^2 26^o + \sin^2 26^o + \frac{\cot 54^o}{\cot 54^o}
\displaystyle = 2
\displaystyle \\
\displaystyle \text{Question 11: } 3 \cos 80^o . \mathrm{cosec} 10^o + 2 \sin 59^o.\sec 31^o \hspace{1.0cm} \text{[ICSE 2013]}
\displaystyle \text{Answer:}
\displaystyle 3 \cos 80^o . \mathrm{cosec} 10^o + 2 \sin 59^o.\sec 31^o
\displaystyle = 3 \cos 80^o . \mathrm{cosec} (90^o - 80^o) + 2 \sin 59^o.\sec (90^o - 59^o)
\displaystyle = 3 \cos 80^o . \sec 80^o + 2 \sin 59^o . \mathrm{cosec} 59^o
\displaystyle = 3 + 2 = 5
\displaystyle \\
\displaystyle \text{Question 12: } \frac{\sin 80^o}{\cos 10^o} + \sin 59^o . \sec 31^o \hspace{1.0cm} \text{[ICSE 2007]}
\displaystyle \text{Answer:}
\displaystyle \frac{\sin 80^o}{\cos 10^o} + \sin 59^o . \sec 31^o
\displaystyle = \frac{\sin (90^o - 10^o)}{\cos 10^o} + \sin 59^o . \sec (90^o - 59^o)
\displaystyle = \frac{\cos 10^o}{\cos 10^o} + \sin 59^o . \mathrm{\mathrm{cosec} } 59^o = 2
\displaystyle \\
\displaystyle \text{Question 13: } 14 \sin 30^o + 6 \cos 60^o - 5 \tan 45^o \hspace{1.0cm} \text{[ICSE 2004]}
\displaystyle \text{Answer:}
\displaystyle 14 \sin 30^o + 6 \cos 60^o - 5 \tan 45^o
\displaystyle = 14 \sin (90^o - 60^o) + 6 \cos 60^o - 5 \tan 45^o
\displaystyle = 14 \cos 60^o + 6 \cos 60^o - 5 \tan 45^o
\displaystyle = 20 \cos 60^o - 5 \tan 45^o
\displaystyle = 20 \times \frac{1}{2} - 5 \times 1
\displaystyle = 10-5 = 5
\displaystyle \\
Question 14: Evaluate without using trigonometric tables:
\displaystyle 2 (\frac{\tan 35^o}{\cot 55^o})^2 + (\frac{\cot 55^o}{\tan 35^o})^2 - 3 (\frac{\sec 40^o}{\mathrm{cosec} 50^o}) \hspace{1.0cm} \text{[ICSE 2011]}
\displaystyle \text{Answer:}
\displaystyle 2 (\frac{\tan 35^o}{\cot 55^o})^2 + (\frac{\cot 55^o}{\tan 35^o})^2 - 3 (\frac{\sec 40^o}{\mathrm{cosec} 50^o})
\displaystyle = 2 (\frac{\tan 35^o}{\cot (90^o - 35^o)})^2 + (\frac{\cot 55^o}{\tan (90^o - 55^o)})^2 - 3 (\frac{sec 40^o}{\mathrm{cosec} (90^o - 40^o)})
\displaystyle = 2 (\frac{\tan 35^o}{\tan 35^o})^2 + (\frac{\cot 55^o}{\cot 55^o})^2 - 3 (\frac{\sec 40^o}{\sec 40^o})
\displaystyle = 2 + 1 - 3 = 0
\displaystyle \\
\displaystyle \text{Question 15: Prove that } (\mathrm{cosec} A- \sin A) (\sec A-\cos A) \sec^2 A = \tan A \hspace{1.0cm} \text{[ICSE 2011]}
\displaystyle \text{Answer:}
\displaystyle (\mathrm{cosec} A- \sin A) (\sec A-\cos A) \sec^2 A = \tan A
\displaystyle \text{LHS } = (\mathrm{cosec} A- \sin A) (\sec A-\cos A) \sec^2 A
\displaystyle = 2 \frac{1 - \sin^2 A}{\sin A} . \frac{1 - \cos^2 A}{\cos A} . \frac{1}{\cos^2 A}
\displaystyle = 2 \frac{\cos^2 A}{\sin A}. \frac{\sin^2 A}{\cos A} . \frac{1}{\cos^2 A}
\displaystyle = 2 \frac{\sin A}{\cos A} =\tan A = \text{ RHS. Hence Proved.}
\displaystyle \\
Question 16: Without using trigonometric tables, evaluate
\displaystyle \sin^2 34^o + \sin^2 56^o + 2 \tan 18^o \tan 72^o - \cot ^2 30^o \hspace{1.0cm} \text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given } \sin^2 34^o + \sin^2 56^o + 2 \tan 18^o \tan 72^o - \cot ^2 30^o
\displaystyle = \sin^2 34^o + \sin^2 (90^o - 34^o) + 2 \tan 18^o \tan (90^o-18^o) - \cot ^2 30^o
\displaystyle = \sin^2 34^o + \cos^2 34^o + 2 \tan 18^o \cot 18^o - (\sqrt{3})^2
\displaystyle = 1+ 2 \tan 18^o \times \frac{1}{\tan 18^o}-3
\displaystyle = 1+2-3 = 0
\displaystyle \\
Question 17: Prove the identity:
\displaystyle (\sin \theta + \cos \theta) (\tan \theta + \cot \theta) = \sec \theta + \mathrm{cosec} \theta \hspace{1.0cm} \text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS } = (\sin \theta + \cos \theta)(\tan \theta+ \cot \theta)
\displaystyle = (\sin \theta + \cos \theta) (\frac{\sin \theta}{\cos \theta} + \frac{cos \theta}{\sin \theta})
\displaystyle = (\sin \theta + \cos \theta) (\frac{\sin^2 \theta+ \cos^2 \theta}{\sin \theta . \cos \theta})
\displaystyle = (\sin \theta + \cos \theta) (\frac{1}{\sin \theta . \cos \theta})
\displaystyle = (\frac{\sin \theta}{\sin \theta . \cos \theta}) + (\frac{\cos \theta}{sin \theta . \cos \theta})
\displaystyle = \frac{1}{\cos \theta}+\frac{1}{\sin \theta}
\displaystyle = \sec \theta+\mathrm{cosec} \theta = \text{ RHS. Hence Proved.}
\displaystyle \\
Question 18: If \displaystyle 2 \sin A - 1 = 0 , show that \displaystyle \sin 3A = 2 \sin A - 4 \sin^3 A \hspace{1.0cm} \text{[ICSE 2001]}
\displaystyle \text{Answer:}
\displaystyle 2 \sin A - 1 = 0
\displaystyle \Rightarrow \sin A = \frac{1}{2} \Rightarrow A = 30^o
\displaystyle \text{Therefore to prove: } \sin 3A = 2 \sin A - 4 \sin^3 A
\displaystyle \text{LHS } = \sin 3A = \sin 90^o = 1
\displaystyle \text{RHS } = 2 \sin A - 4 \sin^3 A = 2 \sin 30^o - 4 \sin^3 30^o = \frac{3}{2} - \frac{1}{2} = 1
Therefore LHS = RHS. Hence proved.
\displaystyle \\
\text{Question 19: Evaluate: } \displaystyle \frac{3 \sin 72^o}{\cos 18^o}- \frac{\sec 32^o}{\mathrm{cosec} 58^o} \hspace{1.0cm} \text{[ICSE 2000]}
\displaystyle \text{Answer:}
\displaystyle \frac{3 \sin 72^o}{\cos 18^o}- \frac{\sec 32^o}{\mathrm{cosec} 58^o}
\displaystyle = \frac{3 \sin 72^o}{\cos (90^o - 72^o)}- \frac{\sec 32^o}{\mathrm{cosec} (90^o - 32^o)}
\displaystyle = \frac{3 \sin 72^o}{\sin 72^o}- \frac{\sec 32^o}{\sec 32^o}
\displaystyle = 3 - 1 = 2
\displaystyle \\
\displaystyle \text{Question 20: Evaluate: } 3 \cos 80^o \mathrm{cosec} 10^o + 2 \cos 59^o \mathrm{cosec} 31^o \hspace{1.0cm} \text{[ICSE 2002]}
\displaystyle \text{Answer:}
\displaystyle 3 \cos 80^o \mathrm{cosec} 10^o + 2 \cos 59^o \mathrm{cosec} 31^o
\displaystyle = 3 \cos 80^o \mathrm{cosec} (90^o - 80^o) + 2 \cos 59^o \mathrm{cosec} (90^o - 59^o)
\displaystyle = 3 \cos 80^o \sec 80^o + 2 \cos 59^o \sec 59^o
\displaystyle = 3 + 2 = 5
\displaystyle \\
\displaystyle \text{Question 21: } \frac{\cos 75^o}{\sin 15^o} + \frac{\sin 12^o}{\cos 78^o} - \frac{\cos 18^o}{\sin 72^o} \hspace{1.0cm} \text{[ICSE 2003]}
\displaystyle \text{Answer:}
\displaystyle \frac{\cos 75^o}{\sin 15^o} + \frac{\sin 12^o}{\cos 78^o} - \frac{\cos 18^o}{\sin 72^o}
\displaystyle = \frac{\cos 75^o}{\sin (90^o - 75^o)} + \frac{\sin 12^o}{\cos (90^o - 12^o)} - \frac{\cos 18^o}{\sin (90^o - 18^o)}
\displaystyle = \frac{\cos 75^o}{\cos 75^o} + \frac{\sin 12^o}{\sin 12^o} - \frac{\cos 18^o}{\cos 18^o}
\displaystyle = 1+ 1 - 1 = 1
\displaystyle \\
\displaystyle \text{Question 22: Prove that: } \frac{\sin A}{1 + \cos A} + \frac{1 +\cos A}{\sin A} = 2 \mathrm{cosec} A \hspace{1.0cm} \text{[ICSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS } = \displaystyle \frac{\sin A}{1 + \cos A} + \frac{1 +\cos A}{\sin A}
\displaystyle = \displaystyle \frac{\sin^2 A + (1 + \cos A)^2}{(1+ \cos A)\sin A}
\displaystyle = \displaystyle \frac{\sin^2 A + 1 + \cos^2 A + 2 \cos A}{(1+ \cos A)\sin A}
\displaystyle = \displaystyle \frac{2(1+\cos A)}{(1+ \cos A)\sin A}
\displaystyle = \displaystyle \frac{2}{\sin A}
\displaystyle = \displaystyle 2 \mathrm{cosec} A = \text{ RHS. Hence Proved.}
\displaystyle \\
\displaystyle \text{Question 23: Prove that: } \frac{\cos A \cot A}{1 - \sin A} = 1 + \mathrm{cosec} A \hspace{1.0cm} \text{[ICSE 2006]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS } = \frac{\cos A. \cot A}{1 - \sin A}
\displaystyle = \frac{\cos A . \frac{\cos A}{\sin A}}{1 - \sin A}
\displaystyle = \frac{\cos^2 A}{\sin A(1 - \sin A)}
\displaystyle = \frac{(1-\sin A)(1 + \sin A)}{\sin A(1 - \sin A)}
\displaystyle = \frac{1 + \sin A}{\sin A}
\displaystyle = \mathrm{cosec} A + 1 = \text{ RHS. Hence Proved.}
\displaystyle \\
Question 24: Without using trigonometric tables evaluate :
\displaystyle \frac{\sin 35^o \cos 55^o + \cos 35^o \sin 55^o}{\mathrm{cosec} ^2 10^o - \tan^2 80^o} \hspace{1.0cm} \text{[ICSE 2010]}
\displaystyle \text{Answer:}
Given: \displaystyle \frac{\sin 35^o \cos 55^o + \cos 35^o \sin 55^o}{\mathrm{cosec} ^2 10^o - \tan^2 80^o}
\displaystyle = \frac{\sin 35^o \cos (90^o- 35^o) + \cos 35^o \sin (90^o- 35^o)}{\mathrm{cosec} ^2 10^o - \tan^2 (90^o - 10^o)}
\displaystyle = \frac{\sin^2 35^o+\cos^2 35^o}{1+ \cot^2 10^o - \cot^2 10^o}
\displaystyle = \frac{1}{1}
\displaystyle = 1
\displaystyle \\
Question 25: Without using trigonometric tables evaluate :
\displaystyle \frac{\sin 35^o \cos 55^o + \cos 35^o \sin 55^o}{\mathrm{cosec} ^2 10^o - \tan^2 80^o} \hspace{1.0cm} \text{[ICSE 2010]}
\displaystyle \text{Answer:}
Given: \displaystyle \frac{\sin 35^o \cos 55^o + \cos 35^o \sin 55^o}{\mathrm{cosec} ^2 10^o - \tan^2 80^o}
\displaystyle = \frac{\sin 35^o \cos (90^o- 35^o) + \cos 35^o \sin (90^o- 35^o)}{\mathrm{cosec} ^2 10^o - \tan^2 (90^o - 10^o)}
\displaystyle = \frac{\sin^2 35^o+\cos^2 35^o}{1+ \cot^2 10^o - \cot^2 10^o}
\displaystyle = \frac{1}{1} = 1

\displaystyle \textbf{Question 26: }~\text{Prove that }1-\frac{\cos^2\theta}{1+\sin\theta}=\sin\theta. \hspace{3.0cm} \text{[ICSE 2001]}
\displaystyle \text{Answer:}
\displaystyle \text{To prove, }1-\frac{\cos^2\theta}{1+\sin\theta}=\sin\theta
\displaystyle \text{LHS}=1-\frac{\cos^2\theta}{1+\sin\theta}
\displaystyle =\frac{1+\sin\theta-\cos^2\theta}{1+\sin\theta}
\displaystyle =\frac{\sin\theta+(1-\cos^2\theta)}{1+\sin\theta}=\frac{\sin\theta+\sin^2\theta}{1+\sin\theta}
\displaystyle \qquad[\text{using identity, }\sin^2A+\cos^2A=1\Rightarrow 1-\cos^2A=\sin^2A]
\displaystyle =\frac{\sin\theta(1+\sin\theta)}{1+\sin\theta}=\sin\theta=\text{RHS}\qquad\text{Hence proved.}

\displaystyle \textbf{Question 27: }~\text{Evaluate without using trigonometric table, } \hspace{3.0cm} \text{[ICSE 2012]}
\displaystyle \cos^2 26^\circ+\cos 64^\circ\sin 26^\circ+\frac{\tan 36^\circ}{\cot 54^\circ}.
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\cos^2 26^\circ+\cos 64^\circ\sin 26^\circ+ \frac{\tan 36^\circ}{\cot 54^\circ}
\displaystyle =\cos^2 26^\circ+\cos(90^\circ-26^\circ)\sin 26^\circ+ \frac{\tan(90^\circ-54^\circ)}{\cot 54^\circ}
\displaystyle =\cos^2 26^\circ+\sin 26^\circ\sin 26^\circ+ \frac{\cot 54^\circ}{\cot 54^\circ}
\displaystyle \qquad[\because\ \tan(90^\circ-\theta)=\cot\theta\ \text{and}\ \cos(90^\circ-\theta)=\sin\theta]
\displaystyle =(\cos^2 26^\circ+\sin^2 26^\circ)+1
\displaystyle =1+1=2\qquad[\because\ \sin^2\theta+\cos^2\theta=1]

\displaystyle \textbf{Question 28: }~\text{Prove that }\frac{\tan^2\theta}{(\sec\theta-1)^2}=\frac{1+\cos\theta}{1-\cos\theta}. \hspace{3.0cm} \text{[ICSE 2012] }
\displaystyle \text{Answer:}
\displaystyle \text{To prove, }\frac{\tan^2\theta}{(\sec\theta-1)^2}=\frac{1+\cos\theta}{1-\cos\theta}
\displaystyle \text{LHS}=\frac{\tan^2\theta}{(\sec\theta-1)^2}=\frac{\sec^2\theta-1}{(\sec\theta-1)^2}
\displaystyle \qquad[\because\ \sec^2A-1=\tan^2A]
\displaystyle =\frac{(\sec\theta-1)(\sec\theta+1)}{(\sec\theta-1)^2}\qquad [\because\ a^2-b^2=(a-b)(a+b)]
\displaystyle =\frac{\sec\theta+1}{\sec\theta-1}
\displaystyle =\frac{\frac{1}{\cos\theta}+1}{\frac{1}{\cos\theta}-1}\qquad [\because\ \sec A=\frac{1}{\cos A}]
\displaystyle =\frac{1+\cos\theta}{1-\cos\theta}=\text{RHS}\qquad\text{Hence proved.}

\displaystyle \textbf{Question 29: }~\text{Show that }\sqrt{\frac{1-\cos A}{1+\cos A}}=\frac{\sin A}{1+\cos A}. \hspace{3.0cm} \text{[ICSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{To show, }\sqrt{\frac{1-\cos A}{1+\cos A}}=\frac{\sin A}{1+\cos A}
\displaystyle \text{LHS}=\sqrt{\frac{1-\cos A}{1+\cos A}\times\frac{1+\cos A}{1+\cos A}}
\displaystyle \qquad[\text{multiplying numerator and denominator by }\sqrt{1+\cos A}]
\displaystyle =\sqrt{\frac{(1-\cos A)(1+\cos A)}{(1+\cos A)^2}}
\displaystyle =\sqrt{\frac{1-\cos^2A}{(1+\cos A)^2}}\qquad[\because\ (a-b)(a+b)=a^2-b^2]
\displaystyle =\sqrt{\frac{\sin^2A}{(1+\cos A)^2}}\qquad[\because\ \sin^2A+\cos^2A=1]
\displaystyle =\frac{\sin A}{1+\cos A}=\text{RHS}\qquad\text{Hence proved.}

\displaystyle \textbf{Question 30: }~\text{Without using trigonometric table, evaluate } \hspace{3.0cm} \text{[ICSE 2016]}
\displaystyle \mathrm{cosec}^2 57^\circ-\tan^2 33^\circ+\cos 44^\circ\cos 46^\circ-\sqrt{2}\cos 45^\circ-\tan^2 60^\circ.
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\mathrm{cosec}^2 57^\circ-\tan^2 33^\circ+\cos 44^\circ\mathrm{cosec} 46^\circ-\sqrt{2}\cos 45^\circ-\tan^2 60^\circ
\displaystyle =\mathrm{cosec}^2(90^\circ-33^\circ)-\tan^2 33^\circ+\sin 46^\circ\mathrm{cosec} 46^\circ-\sqrt{2}\cos 45^\circ-\tan^2 60^\circ
\displaystyle =\sec^2 33^\circ-\tan^2 33^\circ+1-\sqrt{2}\cos 45^\circ-3
\displaystyle \qquad[\because\ \mathrm{cosec}(90^\circ-\theta)=\sec\theta,\ \cos(90^\circ-\theta)=\sin\theta]
\displaystyle =1+1-\sqrt{2}\left(\frac{\sqrt{2}}{2}\right)-3\qquad[\because\ \sec^2\theta-\tan^2\theta=1]
\displaystyle =2-1-3=-2

\displaystyle \textbf{Question 31: }~\text{Prove that }\frac{\sin\theta-2\sin^3\theta}{2\cos^3\theta-\cos\theta}=\tan\theta. \hspace{3.0cm} \text{[ICSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\sin\theta-2\sin^3\theta}{2\cos^3\theta-\cos\theta}
\displaystyle =\frac{\sin\theta(1-2\sin^2\theta)}{\cos\theta(2\cos^2\theta-1)}
\displaystyle =\frac{\sin\theta(1-2(1-\cos^2\theta))}{\cos\theta(2\cos^2\theta-1)}\qquad[\because\ 1-\cos^2A=\sin^2A]
\displaystyle =\frac{\sin\theta(2\cos^2\theta-1)}{\cos\theta(2\cos^2\theta-1)}
\displaystyle =\frac{\sin\theta}{\cos\theta}=\tan\theta=\text{RHS}\qquad\text{Hence proved.}

\displaystyle \textbf{Question 32: }~\text{Prove that }(1+\cot\theta-\mathrm{cosec}\theta)(1+\tan\theta+\sec\theta)=2. \hspace{3.0cm} \text{[ICSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(1+\cot\theta-\mathrm{cosec}\theta)(1+\tan\theta+\sec\theta)
\displaystyle =\left(1+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta}\right)\left(1+\frac{\sin\theta}{\cos\theta}+\frac{1}{\cos\theta}\right)
\displaystyle =\left(\frac{\sin\theta+\cos\theta-1}{\sin\theta}\right)\left(\frac{\cos\theta+\sin\theta+1}{\cos\theta}\right)
\displaystyle =\frac{(\sin\theta+\cos\theta-1)(\sin\theta+\cos\theta+1)}{\sin\theta\cos\theta}
\displaystyle =\frac{(\sin\theta+\cos\theta)^2-1}{\sin\theta\cos\theta}
\displaystyle =\frac{\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta-1}{\sin\theta\cos\theta}
\displaystyle =\frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta}=2=\text{RHS}\qquad\text{Hence proved.}

\displaystyle \textbf{Question 33: }~\text{Prove that }\frac{\cos A}{1+\sin A}+\tan A=\sec A. \hspace{3.0cm} \text{[ICSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\cos A}{1+\sin A}+\tan A
\displaystyle =\frac{\cos A}{1+\sin A}+\frac{\sin A}{\cos A}\qquad[\because\ \tan\theta=\frac{\sin\theta}{\cos\theta}]
\displaystyle =\frac{\cos^2A+\sin A(1+\sin A)}{\cos A(1+\sin A)}
\displaystyle =\frac{\cos^2A+\sin A+\sin^2A}{\cos A(1+\sin A)}
\displaystyle =\frac{1+\sin A}{\cos A(1+\sin A)}\qquad[\because\ \cos^2A+\sin^2A=1]
\displaystyle =\frac{1}{\cos A}=\sec A=\text{RHS}\qquad\text{Hence proved.}

\displaystyle \textbf{Question 34: }~\text{Evaluate without using trigonometric table, } \hspace{3.0cm} \text{[ICSE 2017]}
\displaystyle \sin^2 28^\circ+\sin^2 62^\circ+\tan^2 38^\circ-\cot^2 52^\circ+\frac{1}{4}\sec^2 30^\circ.
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\sin^2 28^\circ+\sin^2 62^\circ+\tan^2 38^\circ-\cot^2 52^\circ+\frac{1}{4}\sec^2 30^\circ
\displaystyle =\sin^2 28^\circ+\sin^2(90^\circ-28^\circ)+\tan^2(90^\circ-52^\circ)-\cot^2 52^\circ+\frac{1}{4}\sec^2 30^\circ
\displaystyle =\sin^2 28^\circ+\cos^2 28^\circ+\cot^2 52^\circ-\cot^2 52^\circ+\frac{1}{4}\sec^2 30^\circ
\displaystyle \qquad[\because\ \sin(90^\circ-\theta)=\cos\theta\ \text{and}\ \tan(90^\circ-\theta)=\cot\theta]
\displaystyle =1+0+\frac{1}{4}\sec^2 30^\circ\qquad[\because\ \sin^2A+\cos^2A=1]
\displaystyle =1+\frac{1}{4}\left(\frac{1}{\cos 30^\circ}\right)^2=1+\frac{1}{4}\left(\frac{2}{\sqrt{3}}\right)^2
\displaystyle =1+\frac{1}{4}\times\frac{4}{3}=1+\frac{1}{3}=\frac{4}{3}

\displaystyle \textbf{Question 35: }~\text{Prove that }\frac{1}{\sec x-\tan x}+\frac{1}{\sec x+\tan x}=\frac{2}{\cos x}.
\displaystyle \text{Answer:} \hspace{3.0cm} \text{[ICSE 2017]}
\displaystyle \text{LHS}=\frac{1}{\sec x-\tan x}+\frac{1}{\sec x+\tan x}
\displaystyle =\frac{\sec x+\tan x+\sec x-\tan x}{(\sec x-\tan x)(\sec x+\tan x)}
\displaystyle =\frac{2\sec x}{\sec^2 x-\tan^2 x}\qquad[\because\ (a-b)(a+b)=a^2-b^2]
\displaystyle =\frac{2\sec x}{1}\qquad[\because\ \sec^2\theta-\tan^2\theta=1]
\displaystyle =2\sec x=\frac{2}{\cos x}=\text{RHS}

\displaystyle \textbf{Question 36: }~\text{Prove that }\sqrt{\sec^2\theta+\mathrm{cosec}^2\theta}=\tan\theta+\cot\theta. \hspace{3.0cm} \text{[ICSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sqrt{\sec^2\theta+\mathrm{cosec}^2\theta}
\displaystyle =\sqrt{1+\tan^2\theta+1+\cot^2\theta}
\displaystyle \qquad[\because\ 1+\tan^2A=\sec^2A\ \text{and}\ 1+\cot^2A=\mathrm{cosec}^2A]
\displaystyle =\sqrt{\tan^2\theta+\cot^2\theta+2}
\displaystyle =\sqrt{\tan^2\theta+\cot^2\theta+2\tan\theta\cot\theta}
\displaystyle \qquad[\because\ \tan A\cdot\cot A=1]
\displaystyle =\sqrt{(\tan\theta+\cot\theta)^2}\qquad[\because\ a^2+b^2+2ab=(a+b)^2]
\displaystyle =\tan\theta+\cot\theta=\text{RHS}\qquad\text{Hence proved.}

\displaystyle \textbf{Question 37: }~\text{Prove that }(\mathrm{cosec}\theta-\sin\theta)(\sec\theta-\cos\theta)(\tan\theta+\cot\theta)=1. \hspace{3.0cm} \text{[ICSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{To prove, }(\mathrm{cosec}\theta-\sin\theta)(\sec\theta-\cos\theta)(\tan\theta+\cot\theta)=1
\displaystyle \text{LHS}=\left(\frac{1}{\sin\theta}-\sin\theta\right)\left(\frac{1}{\cos\theta}-\cos\theta\right)\left(\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}\right)
\displaystyle =\left(\frac{1-\sin^2\theta}{\sin\theta}\right)\left(\frac{1-\cos^2\theta}{\cos\theta}\right)\left(\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}\right)
\displaystyle =\left(\frac{\cos^2\theta}{\sin\theta}\right)\left(\frac{\sin^2\theta}{\cos\theta}\right)\left(\frac{1}{\sin\theta\cos\theta}\right)\qquad[\because\ \sin^2\theta+\cos^2\theta=1]
\displaystyle =1=\text{RHS}\qquad\text{Hence proved.}

\displaystyle \textbf{Question 38: }~\text{Prove that }\frac{\sin A}{1+\cot A}-\frac{\cos A}{1+\tan A}=\sin A-\cos A. \hspace{3.0cm} \text{[ICSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\sin A}{1+\cot A}-\frac{\cos A}{1+\tan A}
\displaystyle =\frac{\sin A}{1+\frac{\cos A}{\sin A}}-\frac{\cos A}{1+\frac{\sin A}{\cos A}}
\displaystyle =\frac{\sin^2A}{\sin A+\cos A}-\frac{\cos^2A}{\cos A+\sin A}
\displaystyle =\frac{\sin^2A-\cos^2A}{\sin A+\cos A}
\displaystyle =\frac{(\sin A+\cos A)(\sin A-\cos A)}{\sin A+\cos A}
\displaystyle \qquad[\because\ a^2-b^2=(a+b)(a-b)]
\displaystyle =\sin A-\cos A=\text{RHS}\qquad\text{Hence proved.}

\displaystyle \textbf{Question 39: }~\text{Prove the following identity }(\sin A+\mathrm{cosec}A)^2+(\cos A+\sec A)^2=5+\sec^2A\cdot\mathrm{cosec}^2A. \hspace{3.0cm} \text{[ICSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(\sin A+\mathrm{cosec}A)^2+(\cos A+\sec A)^2
\displaystyle =\sin^2A+\mathrm{cosec}^2A+2\sin A\,\mathrm{cosec}A+\cos^2A+\sec^2A+2\cos A\,\sec A
\displaystyle \qquad[\because\ (a+b)^2=a^2+b^2+2ab]
\displaystyle =(\sin^2A+\cos^2A)+2+2+\sec^2A+\mathrm{cosec}^2A
\displaystyle \qquad\left[\because\ \sec\theta=\frac{1}{\cos\theta}\ \text{and}\ \mathrm{cosec}\theta=\frac{1}{\sin\theta}\right]
\displaystyle =1+4+\sec^2A+\mathrm{cosec}^2A=5+\sec^2A+\mathrm{cosec}^2A
\displaystyle =5+\frac{\sin^2A+\cos^2A}{\cos^2A\sin^2A}
\displaystyle =5+\frac{1}{\cos^2A\sin^2A}\qquad[\because\ \sin^2\theta+\cos^2\theta=1]
\displaystyle =5+\sec^2A\,\mathrm{cosec}^2A=\text{RHS}

\displaystyle \textbf{Question 40: }~\text{Prove the following identity } \hspace{3.0cm} \text{[ICSE 2020]}
\displaystyle \frac{\sec A}{\sec A-1}+\frac{\sec A}{\sec A+1}=2\,\mathrm{cosec}^2A.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\sec A}{\sec A-1}+\frac{\sec A}{\sec A+1}
\displaystyle =\frac{\sec A(\sec A+1)+\sec A(\sec A-1)}{(\sec A-1)(\sec A+1)}
\displaystyle =\frac{\sec A(2\sec A)}{\sec^2A-1}
\displaystyle \qquad[\because\ (a-b)(a+b)=a^2-b^2]
\displaystyle =\frac{2\sec^2A}{\tan^2A}\qquad[\because\ 1+\tan^2\theta=\sec^2\theta]
\displaystyle =2\times\frac{\frac{1}{\cos^2A}}{\frac{\sin^2A}{\cos^2A}}
\displaystyle \qquad[\because\ \tan\theta=\frac{\sin\theta}{\cos\theta}\ \text{and}\ \sec\theta=\frac{1}{\cos\theta}]
\displaystyle =\frac{2}{\sin^2A}=2\,\mathrm{cosec}^2A

\displaystyle \textbf{Question 41: }~\text{Prove that }\frac{1}{1+\sin\theta}+\frac{1}{1-\sin\theta}=2\sec^2\theta. \hspace{3.0cm} \text{[ICSE 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{1}{1+\sin\theta}+\frac{1}{1-\sin\theta}
\displaystyle =\frac{1-\sin\theta+1+\sin\theta}{1-\sin^2\theta}
\displaystyle =\frac{2}{\cos^2\theta}\qquad[\because\ 1-\sin^2A=\cos^2A]
\displaystyle =2\sec^2\theta=\text{RHS}\qquad\text{Hence proved.}

\displaystyle \textbf{Question 42: }~\text{Prove that }1+\frac{\tan^2\theta}{1+\sec\theta}=\sec\theta. \hspace{3.0cm} \text{[ICSE 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=1+\frac{\tan^2\theta}{1+\sec\theta}=1+\frac{\sec^2\theta-1}{1+\sec\theta}
\displaystyle \qquad[\because\ \tan^2A=\sec^2A-1]
\displaystyle =1+\frac{(\sec\theta+1)(\sec\theta-1)}{1+\sec\theta}
\displaystyle \qquad[\because\ a^2-b^2=(a+b)(a-b)]
\displaystyle =1+\sec\theta-1
\displaystyle =\sec\theta=\text{RHS}\qquad\text{Hence proved.}

\displaystyle \textbf{Question 43: }~\tan\theta\left(\sqrt{1-\sin^2\theta}\right)\text{ is equal to}
\displaystyle \text{(a) }\cos\theta \quad \text{(b) }\sin\theta \quad \text{(c) }\tan\theta \quad \text{(d) }\cot\theta \hspace{3.0cm} \text{[ICSE 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{(b) We have, }\tan\theta\times\sqrt{1-\sin^2\theta}
\displaystyle =\tan\theta\times\sqrt{\cos^2\theta}\qquad[\because\ 1-\sin^2A=\cos^2A]
\displaystyle =\tan\theta\cdot\cos\theta=\sin\theta

\displaystyle \textbf{Question 44: }~\text{Prove that }\frac{(1+\sin\theta)^2+(1-\sin\theta)^2}{2\cos^2\theta}=\sec^2\theta+\tan^2\theta. \hspace{3.0cm} \text{[ICSE 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{(1+\sin\theta)^2+(1-\sin\theta)^2}{2\cos^2\theta}
\displaystyle =\frac{1+\sin^2\theta+2\sin\theta+1+\sin^2\theta-2\sin\theta}{2\cos^2\theta}
\displaystyle \qquad[\because\ (a\pm b)^2=a^2+b^2\pm2ab]
\displaystyle =\frac{2(1+\sin^2\theta)}{2\cos^2\theta}=\frac{1+\sin^2\theta}{\cos^2\theta}
\displaystyle =\frac{1}{\cos^2\theta}+\frac{\sin^2\theta}{\cos^2\theta}=\sec^2\theta+\tan^2\theta=\text{RHS}
\displaystyle \qquad\left[\because\ \frac{1}{\cos A}=\sec A\ \text{and}\ \tan A=\frac{\sin A}{\cos A}\right]
\displaystyle \text{Hence proved.}

\displaystyle \textbf{Question 45: }~(1+\sin A)(1-\sin A)\text{ is equal to}
\displaystyle \text{(a) }\tan^2A \quad \text{(b) }\sin^2A \quad \text{(c) }\sec^2A \quad \text{(d) }\cos^2A \hspace{3.0cm} \text{[ICSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{(d) Given, }(1+\sin A)(1-\sin A)
\displaystyle =1^2-\sin^2A\qquad[\because\ a^2-b^2=(a+b)(a-b)]
\displaystyle =1-\sin^2A
\displaystyle \qquad[\because\ \sin^2A+\cos^2A=1\Rightarrow 1-\sin^2A=\cos^2A]
\displaystyle =\cos^2A

\displaystyle \textbf{Question 46: }~\text{Without using trigonometric table, evaluate } \hspace{3.0cm} \text{[ICSE 2014] }
\displaystyle \sin^2 34^\circ+\sin^2 56^\circ+2\tan 18^\circ\tan 72^\circ-\cot^2 30^\circ.
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\sin^2 34^\circ+\sin^2 56^\circ+2\tan 18^\circ\tan 72^\circ-\cot^2 30^\circ
\displaystyle =\sin^2 34^\circ+\sin^2(90^\circ-34^\circ)+2\tan 18^\circ\tan(90^\circ-18^\circ)-\cot^2 30^\circ
\displaystyle =\sin^2 34^\circ+\cos^2 34^\circ+2\tan 18^\circ\cot 18^\circ-(\sqrt{3})^2
\displaystyle \qquad[\because\ \sin(90^\circ-\theta)=\cos\theta\ \text{and}\ \tan(90^\circ-\theta)=\cot\theta]
\displaystyle =1+2\times1-3\qquad[\because\ \sin^2A+\cos^2A=1\ \text{and}\ \tan\theta\cot\theta=1]
\displaystyle =3-3=0

\displaystyle \textbf{Question 47: }~\text{Without using trigonometric table, evaluate } \hspace{3.0cm} \text{[ICSE 2008]}
\displaystyle \frac{\sin 25^\circ}{\sec 65^\circ}+\frac{\cos 25^\circ}{\mathrm{cosec} 65^\circ}.
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\frac{\sin 25^\circ}{\sec 65^\circ}+ \frac{\cos 25^\circ}{\mathrm{cosec} 65^\circ}
\displaystyle =\frac{\sin 25^\circ}{\sec(90^\circ-25^\circ)}+ \frac{\cos 25^\circ}{\mathrm{cosec}(90^\circ-25^\circ)}
\displaystyle =\frac{\sin 25^\circ}{\mathrm{cosec} 25^\circ}+ \frac{\cos 25^\circ}{\sec 25^\circ}
\displaystyle \qquad[\because\ \sec(90^\circ-\theta)=\mathrm{cosec}\theta\ \text{and}\ \mathrm{cosec}(90^\circ-\theta)=\sec\theta]
\displaystyle =\sin^2 25^\circ+\cos^2 25^\circ\qquad \left[\because\ \mathrm{cosec}\theta=\frac{1}{\sin\theta}\ \text{and}\ \sec\theta=\frac{1}{\cos\theta}\right]
\displaystyle =1\qquad[\text{using identity, }\sin^2\theta+\cos^2\theta=1]

\displaystyle \textbf{Question 48: }~\text{Prove the identity }
\displaystyle \frac{\sin A}{1+\cos A}=\mathrm{cosec}A-\cot A. \hspace{3.0cm} \text{[ICSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{To prove, }\frac{\sin A}{1+\cos A}=\mathrm{cosec}A-\cot A
\displaystyle \text{LHS}=\frac{\sin A}{1+\cos A}\times\frac{1-\cos A}{1-\cos A}
\displaystyle \qquad[\text{multiplying numerator and denominator by }(1-\cos A)]
\displaystyle =\frac{\sin A(1-\cos A)}{1^2-\cos^2A}\qquad [\because\ (a-b)(a+b)=a^2-b^2]
\displaystyle =\frac{\sin A(1-\cos A)}{\sin^2A}\qquad [\text{using identity, }\sin^2\theta+\cos^2\theta=1\Rightarrow \sin^2\theta=1-\cos^2\theta]
\displaystyle =\frac{1-\cos A}{\sin A}=\frac{1}{\sin A}-\frac{\cos A}{\sin A}
\displaystyle =\mathrm{cosec}A-\cot A=\text{RHS}\qquad \left[\because\ \cot\theta=\frac{\cos\theta}{\sin\theta}\ \text{and}\ \frac{1}{\sin\theta}=\mathrm{cosec}\theta\right]
\displaystyle \text{Hence proved.}

\displaystyle \textbf{Question 49: }~\text{Factorise }\sin^3\theta+\cos^3\theta. \hspace{3.0cm} \text{[ICSE 2008]}
\displaystyle \text{Hence, prove the identity }\frac{\sin^3\theta+\cos^3\theta}{\sin\theta+\cos\theta}+\sin\theta\cos\theta=1.
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\sin^3\theta+\cos^3\theta
\displaystyle \qquad[\text{Using the identity }a^3+b^3=(a+b)(a^2+b^2-ab)]
\displaystyle \therefore\ \sin^3\theta+\cos^3\theta=(\sin\theta+\cos\theta)(\sin^2\theta+\cos^2\theta-\sin\theta\cos\theta)
\displaystyle =(\sin\theta+\cos\theta)(1-\sin\theta\cos\theta)
\displaystyle \text{which is the required factor form.}
\displaystyle \therefore\ \frac{\sin^3\theta+\cos^3\theta}{\sin\theta+\cos\theta}=1-\sin\theta\cos\theta
\displaystyle \therefore\ \frac{\sin^3\theta+\cos^3\theta}{\sin\theta+\cos\theta}+\sin\theta\cos\theta=1\qquad\text{Hence proved.}

\displaystyle \textbf{Question 50: }~\text{Without using trigonometric table, evaluate } \hspace{3.0cm} \text{[ICSE 2009]}
\displaystyle \frac{\sec 17^\circ}{\mathrm{cosec} 73^\circ}+\frac{\tan 68^\circ}{\cot 22^\circ}+\cos^2 44^\circ+\cos^2 46^\circ.
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\frac{\sec 17^\circ}{\mathrm{cosec} 73^\circ}+ \frac{\tan 68^\circ}{\cot 22^\circ}+\cos^2 44^\circ+\cos^2 46^\circ
\displaystyle =\frac{\sec(90^\circ-73^\circ)}{\mathrm{cosec} 73^\circ}+ \frac{\tan(90^\circ-22^\circ)}{\cot 22^\circ}+\cos^2 44^\circ+\cos^2(90^\circ-44^\circ)
\displaystyle =\frac{\mathrm{cosec} 73^\circ}{\mathrm{cosec} 73^\circ}+ \frac{\cot 22^\circ}{\cot 22^\circ}+(\cos^2 44^\circ+\sin^2 44^\circ)
\displaystyle \qquad[\because\ \sec(90^\circ-\theta)=\mathrm{cosec}\theta,\ \tan(90^\circ-\theta)=\cot\theta\ \text{and}\ \cos(90^\circ-\theta)=\sin\theta]
\displaystyle =1+1+(1)=3\qquad[\text{using identity, }\sin^2\theta+\cos^2\theta=1]

\displaystyle \textbf{Question 51: }~\text{Prove the following identity } \hspace{3.0cm} \text{[ICSE 2009]}
\displaystyle \frac{\sin A}{1+\cos A}+\frac{1+\cos A}{\sin A}=2\,\mathrm{cosec}A.
\displaystyle \text{Answer:}
\displaystyle \text{To prove, }\frac{\sin A}{1+\cos A}+\frac{1+\cos A}{\sin A}= 2\,\mathrm{cosec}A
\displaystyle \text{LHS}=\frac{\sin A}{1+\cos A}+\frac{1+\cos A}{\sin A}= \frac{\sin^2A+(1+\cos A)^2}{\sin A(1+\cos A)}
\displaystyle =\frac{\sin^2A+1+\cos^2A+2\cos A}{\sin A(1+\cos A)}\qquad [\because\ (a+b)^2=a^2+b^2+2ab]
\displaystyle =\frac{(\sin^2A+\cos^2A)+1+2\cos A}{\sin A(1+\cos A)}\qquad [\text{using identity, }\sin^2\theta+\cos^2\theta=1]
\displaystyle =\frac{2(1+\cos A)}{\sin A(1+\cos A)}=\frac{2}{\sin A}
\displaystyle =2\,\mathrm{cosec}A=\text{RHS}\qquad \left[\because\ \frac{1}{\sin A}=\mathrm{cosec}A\right]
\displaystyle \text{Hence proved.}

\displaystyle \textbf{Question 52: }~\text{Without using trigonometric table, evaluate } \hspace{3.0cm} \text{[ICSE 2010]}
\displaystyle \frac{\sin 35^\circ+\cos 35^\circ\sin 55^\circ}{\mathrm{cosec}^2 10^\circ-\tan^2 80^\circ}.
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\frac{\sin 35^\circ\cos 55^\circ+ \cos 35^\circ\sin 55^\circ}{\mathrm{cosec}^2 10^\circ-\tan^2 80^\circ}
\displaystyle =\frac{\sin(90^\circ-55^\circ)\cos 55^\circ+ \cos(90^\circ-55^\circ)\sin 55^\circ}{\mathrm{cosec}^2 10^\circ-\tan^2(90^\circ-10^\circ)}
\displaystyle =\frac{\cos 55^\circ\cos 55^\circ+\sin 55^\circ\sin 55^\circ} {\mathrm{cosec}^2 10^\circ-\cot^2 10^\circ}
\displaystyle \qquad[\because\ \sin(90^\circ-\theta)=\cos\theta,\ \cos(90^\circ-\theta)=\sin\theta\ \text{and}\ \tan(90^\circ-\theta)=\cot\theta]
\displaystyle =\frac{\cos^2 55^\circ+\sin^2 55^\circ}{1}\qquad [\because\ \mathrm{cosec}^2\theta-\cot^2\theta=1]
\displaystyle =\frac{1}{1}=1\qquad[\because\ \cos^2\theta+\sin^2\theta=1]

\displaystyle \textbf{Question 53: }~\text{Prove that }\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2\sec A. \hspace{3.0cm} \text{[ICSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}
\displaystyle =\frac{\cos^2A+(1+\sin A)^2}{\cos A(1+\sin A)}
\displaystyle =\frac{\cos^2A+1+\sin^2A+2\sin A}{\cos A(1+\sin A)}
\displaystyle =\frac{(\sin^2A+\cos^2A)+1+2\sin A}{\cos A(1+\sin A)}
\displaystyle =\frac{1+1+2\sin A}{\cos A(1+\sin A)}=\frac{2(1+\sin A)}{\cos A(1+\sin A)}
\displaystyle =\frac{2}{\cos A}=2\sec A=\text{RHS}\qquad\text{Hence proved.}

\displaystyle \textbf{Question 54: }~\text{Prove the identity }\left(\frac{1-\tan\theta}{1-\cot\theta}\right)^2=\tan^2\theta. \hspace{3.0cm} \text{[ICSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\left(\frac{1-\tan\theta}{1-\cot\theta}\right)^2
\displaystyle =\left(\frac{1-\tan\theta}{1-\frac{1}{\tan\theta}}\right)^2\qquad[\because\ \cot A=\frac{1}{\tan A}]
\displaystyle =\left(\frac{1-\tan\theta}{\frac{\tan\theta-1}{\tan\theta}}\right)^2
\displaystyle =\left(\frac{\tan\theta(1-\tan\theta)}{\tan\theta-1}\right)^2=\tan^2\theta=\text{RHS}\qquad\text{Hence proved.}

\displaystyle \textbf{Question 55: }~\text{Prove the following identity }(\sin^2\theta-1)(\tan^2\theta+1)+1=0. \hspace{3.0cm} \text{[ICSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(\sin^2\theta-1)(\tan^2\theta+1)+1
\displaystyle \text{Identities: }\sec^2\theta-\tan^2\theta=1\ \text{and}\ \sin^2\theta+\cos^2\theta=1
\displaystyle \Rightarrow\ \sec^2\theta=1+\tan^2\theta
\displaystyle \text{and}\ -\cos^2\theta=\sin^2\theta-1
\displaystyle \text{LHS}=(\sin^2\theta-1)(\tan^2\theta+1)+1
\displaystyle =-\cos^2\theta\sec^2\theta+1
\displaystyle =-\cos^2\theta\left(\frac{1}{\cos^2\theta}\right)+1
\displaystyle =-1+1=0=\text{RHS}

\displaystyle \textbf{Question 56: }~\text{Prove that }\frac{\sin\theta}{1-\cot\theta}+\frac{\cos\theta}{1-\tan\theta}=\cos\theta+\sin\theta. \hspace{3.0cm} \text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\sin\theta}{1-\cot\theta}+\frac{\cos\theta}{1-\tan\theta}
\displaystyle =\frac{\sin\theta}{1-\frac{\cos\theta}{\sin\theta}}+ \frac{\cos\theta}{1-\frac{\sin\theta}{\cos\theta}}\qquad [\because\ \cot A=\frac{\cos A}{\sin A}\ \text{and}\ \tan A=\frac{\sin A}{\cos A}]
\displaystyle =\frac{\sin^2\theta}{\sin\theta-\cos\theta}+ \frac{\cos^2\theta}{\cos\theta-\sin\theta}
\displaystyle =\frac{\sin^2\theta-\cos^2\theta}{\sin\theta-\cos\theta}
\displaystyle =\frac{(\sin\theta+\cos\theta)(\sin\theta-\cos\theta)}{\sin\theta-\cos\theta}\qquad [\because\ a^2-b^2=(a+b)(a-b)]
\displaystyle =\sin\theta+\cos\theta=\text{RHS}\qquad\text{Hence proved.}

\displaystyle \textbf{Question 57: }~\text{Prove the identity }(\sin\theta+\cos\theta)(\tan\theta+\cot\theta)=\sec\theta+\mathrm{cosec}\theta. \hspace{3.0cm} \text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{To prove, }(\sin\theta+\cos\theta)(\tan\theta+\cot\theta)= \sec\theta+\mathrm{cosec}\theta
\displaystyle \text{LHS}=(\sin\theta+\cos\theta) \left(\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}\right)
\displaystyle \qquad[\because\ \tan A=\frac{\sin A}{\cos A}\ \text{and}\ \cot A=\frac{\cos A}{\sin A}]
\displaystyle =(\sin\theta+\cos\theta) \left(\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}\right)
\displaystyle =(\sin\theta+\cos\theta)\frac{1}{\sin\theta\cos\theta} \qquad[\text{using identity, }\sin^2\theta+\cos^2\theta=1]
\displaystyle =\frac{\sin\theta}{\sin\theta\cos\theta}+ \frac{\cos\theta}{\sin\theta\cos\theta}
\displaystyle =\frac{1}{\cos\theta}+\frac{1}{\sin\theta}
\displaystyle =\sec\theta+\mathrm{cosec}\theta=\text{RHS}\qquad \left[\because\ \sec\theta=\frac{1}{\cos\theta}\ \text{and}\ \mathrm{cosec}\theta=\frac{1}{\sin\theta}\right]
\displaystyle \text{Hence proved.}

\displaystyle \textbf{Question 58: }~\text{Prove that }(\mathrm{cosec}A-\sin A)(\sec A-\cos A)\sec^2A=\tan A. \hspace{3.0cm} \text{[ICSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{To prove, }(\mathrm{cosec}A-\sin A) (\sec A-\cos A)\sec^2A=\tan A
\displaystyle \text{LHS}=(\mathrm{cosec}A-\sin A) (\sec A-\cos A)\sec^2A
\displaystyle =\left(\frac{1}{\sin A}-\sin A\right) \left(\frac{1}{\cos A}-\cos A\right)\sec^2A \qquad[\because\ \sec A=\frac{1}{\cos A}\ \text{and}\ \mathrm{cosec}A=\frac{1}{\sin A}]
\displaystyle =\left(\frac{1-\sin^2A}{\sin A}\right) \left(\frac{1-\cos^2A}{\cos A}\right)\sec^2A
\displaystyle =\left(\frac{\cos^2A}{\sin A}\right) \left(\frac{\sin^2A}{\cos A}\right)\frac{1}{\cos^2A} \qquad[\text{using identity, }\sin^2A+\cos^2A=1]
\displaystyle =\frac{\sin A}{\cos A}=\tan A=\text{RHS} \qquad\text{Hence proved.}

\displaystyle \textbf{Question 59: }~\text{Statement I: }\sin^2\theta+\cos^2\theta=1
\displaystyle \text{Statement II: }\mathrm{cosec}^2\theta+\cot^2\theta=1
\displaystyle \text{Which of the following is valid?}
\displaystyle \text{(a) only I \quad (b) only II \quad (c) Both I and II \quad (d) Neither I nor II} \hspace{3.0cm} \text{[ICSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Statement I }\sin^2\theta+\cos^2\theta=1\ \text{which is true.}
\displaystyle \text{Statement II }\mathrm{cosec}^2\theta+\cot^2\theta=1\ \text{which is false.}

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