\displaystyle \textbf{Prove the following identities:}
\displaystyle \textbf{Question 1: }\frac{\sec A-1}{\sec A+1}=\frac{1-\cos A}{1+\cos A}\qquad[\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{\frac{1-\cos A}{\cos A}}{\frac{1+\cos A}{\cos A}}
\displaystyle =\frac{1-\cos A}{1+\cos A}
\displaystyle =\text{RHS}
\displaystyle \therefore\text{ The identity is proved.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\frac{1+\sin A}{1-\sin A}=\frac{\mathrm{cosec}A+1}{\mathrm{cosec}A-1}
\displaystyle \text{Answer:}
\displaystyle \text{RHS}=\frac{\mathrm{cosec}A+1}{\mathrm{cosec}A-1}
\displaystyle =\frac{\frac{1}{\sin A}+1}{\frac{1}{\sin A}-1}
\displaystyle =\frac{\frac{1+\sin A}{\sin A}}{\frac{1-\sin A}{\sin A}}
\displaystyle =\frac{1+\sin A}{1-\sin A}
\displaystyle =\text{LHS}
\displaystyle \therefore\text{ The identity is proved.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\frac{1}{\tan A+\cot A}=\cos A\sin A
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{1}{\tan A+\cot A}
\displaystyle =\frac{1}{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}}
\displaystyle =\frac{1}{\frac{\sin^2A+\cos^2A}{\sin A\cos A}}
\displaystyle =\frac{\sin A\cos A}{\sin^2A+\cos^2A}
\displaystyle =\sin A\cos A
\displaystyle =\text{RHS}
\displaystyle \therefore\text{ The identity is proved.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\tan A-\cot A=\frac{1-2\cos^2A}{\sin A\cos A}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\tan A-\cot A
\displaystyle =\frac{\sin A}{\cos A}-\frac{\cos A}{\sin A}
\displaystyle =\frac{\sin^2A-\cos^2A}{\sin A\cos A}
\displaystyle =\frac{1-\cos^2A-\cos^2A}{\sin A\cos A}
\displaystyle =\frac{1-2\cos^2A}{\sin A\cos A}
\displaystyle =\text{RHS}
\displaystyle \therefore\text{ The identity is proved.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\sin^4A-\cos^4A=2\sin^2A-1
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sin^4A-\cos^4A
\displaystyle =(\sin^2A-\cos^2A)(\sin^2A+\cos^2A)
\displaystyle =\sin^2A-\cos^2A
\displaystyle =1-\cos^2A-\cos^2A
\displaystyle =1-2\cos^2A
\displaystyle =2(1-\cos^2A)-1
\displaystyle =2\sin^2A-1
\displaystyle =\text{RHS}
\displaystyle \therefore\text{ The identity is proved.}
\displaystyle \\

\displaystyle \textbf{Question 6: }(1-\tan A)^2+(1+\tan A)^2=2\sec^2A\qquad[\text{ICSE }2005]
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(1-\tan A)^2+(1+\tan A)^2
\displaystyle =\left(1-\frac{\sin A}{\cos A}\right)^2+\left(1+\frac{\sin A}{\cos A}\right)^2
\displaystyle =\frac{(\cos A-\sin A)^2}{\cos^2A}+\frac{(\cos A+\sin A)^2}{\cos^2A}
\displaystyle =\frac{\cos^2A+\sin^2A-2\sin A\cos A}{\cos^2A}
\displaystyle \quad+\frac{\cos^2A+\sin^2A+2\sin A\cos A}{\cos^2A}
\displaystyle =\frac{2(\sin^2A+\cos^2A)}{\cos^2A}
\displaystyle =\frac{2}{\cos^2A}
\displaystyle =2\sec^2A
\displaystyle =\text{RHS}
\displaystyle \therefore\text{ The identity is proved.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\mathrm{cosec}^4A-\mathrm{cosec}^2A=\cot^4A+\cot^2A
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\mathrm{cosec}^4A-\mathrm{cosec}^2A
\displaystyle =\mathrm{cosec}^2A\left(\mathrm{cosec}^2A-1\right)
\displaystyle =\mathrm{cosec}^2A\cot^2A
\displaystyle =\left(1+\cot^2A\right)\cot^2A
\displaystyle =\cot^2A+\cot^4A
\displaystyle =\cot^4A+\cot^2A
\displaystyle =\text{RHS}
\displaystyle \therefore\text{ The identity is proved.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\sec A(1-\sin A)(\sec A+\tan A)=1
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sec A(1-\sin A)(\sec A+\tan A)
\displaystyle =\frac{1}{\cos A}(1-\sin A)\left(\frac{1+\sin A}{\cos A}\right)
\displaystyle =\frac{(1-\sin A)(1+\sin A)}{\cos^2A}
\displaystyle =\frac{1-\sin^2A}{\cos^2A}
\displaystyle =\frac{\cos^2A}{\cos^2A}
\displaystyle =1
\displaystyle =\text{RHS}
\displaystyle \therefore\text{ The identity is proved.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\mathrm{cosec}A(1+\cos A)(\mathrm{cosec}A-\cot A)=1
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\mathrm{cosec}A(1+\cos A)(\mathrm{cosec}A-\cot A)
\displaystyle =\frac{1}{\sin A}(1+\cos A)\left(\frac{1}{\sin A}-\frac{\cos A}{\sin A}\right)
\displaystyle =\frac{(1+\cos A)(1-\cos A)}{\sin^2A}
\displaystyle =\frac{1-\cos^2A}{\sin^2A}
\displaystyle =\frac{\sin^2A}{\sin^2A}
\displaystyle =1
\displaystyle =\text{RHS}
\displaystyle \therefore\text{ The identity is proved.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\sec^2A+\mathrm{cosec}^2A=\sec^2A\,\mathrm{cosec}^2A
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sec^2A+\mathrm{cosec}^2A
\displaystyle =\frac{1}{\cos^2A}+\frac{1}{\sin^2A}
\displaystyle =\frac{\sin^2A+\cos^2A}{\sin^2A\cos^2A}
\displaystyle =\frac{1}{\sin^2A\cos^2A}
\displaystyle =\frac{1}{\cos^2A}\times\frac{1}{\sin^2A}
\displaystyle =\sec^2A\,\mathrm{cosec}^2A
\displaystyle =\text{RHS}
\displaystyle \therefore\text{ The identity is proved.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\frac{(1+\tan^2A)\cot A}{\mathrm{cosec}^2A}=\tan A.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{(1+\tan^2A)\cot A}{\mathrm{cosec}^2A}
\displaystyle =\frac{\frac{\cos^2A+\sin^2A}{\cos^2A}\cdot\frac{\cos A}{\sin A}}{\frac{1}{\sin^2A}}
\displaystyle =\frac{\cos^2A+\sin^2A}{\cos^2A}\cdot\frac{\cos A}{\sin A}\cdot\sin^2A
\displaystyle =\frac{1}{\cos A}\cdot\sin A
\displaystyle =\frac{\sin A}{\cos A}=\tan A=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\tan^2A-\sin^2A=\tan^2A\cdot\sin^2A.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\tan^2A-\sin^2A
\displaystyle =\frac{\sin^2A}{\cos^2A}-\sin^2A
\displaystyle =\frac{\sin^2A-\sin^2A\cos^2A}{\cos^2A}
\displaystyle =\frac{\sin^2A(1-\cos^2A)}{\cos^2A}
\displaystyle =\frac{\sin^4A}{\cos^2A}
\displaystyle =\tan^2A\cdot\sin^2A=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\cot^2A-\cos^2A=\cos^2A\cdot\cot^2A.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\cot^2A-\cos^2A
\displaystyle =\frac{\cos^2A}{\sin^2A}-\cos^2A
\displaystyle =\frac{\cos^2A-\cos^2A\sin^2A}{\sin^2A}
\displaystyle =\frac{\cos^2A(1-\sin^2A)}{\sin^2A}
\displaystyle =\frac{\cos^4A}{\sin^2A}
\displaystyle =\cot^2A\cdot\cos^2A=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 14: }(\mathrm{cosec}A+\sin A)(\mathrm{cosec}A-\sin A)=\cot^2A+\cos^2A.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(\mathrm{cosec}A+\sin A)(\mathrm{cosec}A-\sin A)
\displaystyle =\left(\frac{1}{\sin A}+\sin A\right)\left(\frac{1}{\sin A}-\sin A\right)
\displaystyle =\frac{1}{\sin^2A}-\sin^2A
\displaystyle =\frac{1-\sin^4A}{\sin^2A}
\displaystyle =\frac{(1-\sin^2A)(1+\sin^2A)}{\sin^2A}
\displaystyle =\frac{\cos^2A(1+\sin^2A)}{\sin^2A}
\displaystyle =\frac{\cos^2A}{\sin^2A}+\cos^2A
\displaystyle =\cot^2A+\cos^2A=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 15: }(\sec A-\cos A)(\sec A+\cos A)=\sin^2A+\tan^2A.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(\sec A-\cos A)(\sec A+\cos A)
\displaystyle =\left(\frac{1}{\cos A}-\cos A\right)\left(\frac{1}{\cos A}+\cos A\right)
\displaystyle =\frac{1}{\cos^2A}-\cos^2A
\displaystyle =\frac{1-\cos^4A}{\cos^2A}
\displaystyle =\frac{(1-\cos^2A)(1+\cos^2A)}{\cos^2A}
\displaystyle =\frac{\sin^2A(1+\cos^2A)}{\cos^2A}
\displaystyle =\frac{\sin^2A}{\cos^2A}+\sin^2A
\displaystyle =\tan^2A+\sin^2A=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 16: }(\cos A+\sin A)^2+(\cos A-\sin A)^2=2.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(\cos A+\sin A)^2+(\cos A-\sin A)^2
\displaystyle =\cos^2A+\sin^2A+2\cos A\sin A+\cos^2A+\sin^2A-2\cos A\sin A
\displaystyle =2(\cos^2A+\sin^2A)
\displaystyle =2(1)=2=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 17: }(\mathrm{cosec}A-\sin A)(\sec A-\cos A)(\tan A+\cot A)=1.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(\mathrm{cosec}A-\sin A)(\sec A-\cos A)(\tan A+\cot A)
\displaystyle =\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right)\left(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right)
\displaystyle =\frac{1-\sin^2A}{\sin A}\cdot\frac{1-\cos^2A}{\cos A}\cdot\frac{\sin^2A+\cos^2A}{\sin A\cos A}
\displaystyle =\frac{\cos^2A}{\sin A}\cdot\frac{\sin^2A}{\cos A}\cdot\frac{1}{\sin A\cos A}
\displaystyle =1=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 18: }\frac{1}{\sec A+\tan A}=\sec A-\tan A.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{1}{\sec A+\tan A}
\displaystyle =\frac{1}{\frac{1}{\cos A}+\frac{\sin A}{\cos A}}
\displaystyle =\frac{\cos A}{1+\sin A}
\displaystyle =\frac{\cos A}{1+\sin A}\cdot\frac{1-\sin A}{1-\sin A}
\displaystyle =\frac{\cos A(1-\sin A)}{1-\sin^2A}
\displaystyle =\frac{\cos A(1-\sin A)}{\cos^2A}
\displaystyle =\frac{1-\sin A}{\cos A}
\displaystyle =\frac{1}{\cos A}-\frac{\sin A}{\cos A}
\displaystyle =\sec A-\tan A=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 19: }\mathrm{cosec}A+\cot A=\frac{1}{\mathrm{cosec}A-\cot A}.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\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}\cdot\frac{1-\cos A}{1-\cos A}
\displaystyle =\frac{1-\cos^2A}{\sin A(1-\cos A)}
\displaystyle =\frac{\sin^2A}{\sin A(1-\cos A)}
\displaystyle =\frac{\sin A}{1-\cos A}
\displaystyle \text{RHS}=\frac{1}{\mathrm{cosec}A-\cot A}
\displaystyle =\frac{1}{\frac{1}{\sin A}-\frac{\cos A}{\sin A}}
\displaystyle =\frac{1}{\frac{1-\cos A}{\sin A}}
\displaystyle =\frac{\sin A}{1-\cos A}
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 20: }\frac{\sec A-\tan A}{\sec A+\tan A}=2\sec^2A-1-2\sec A\tan A.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\sec A-\tan A}{\sec A+\tan A}
\displaystyle =\frac{\frac{1}{\cos A}-\frac{\sin A}{\cos A}}{\frac{1}{\cos A}+\frac{\sin A}{\cos A}}
\displaystyle =\frac{1-\sin A}{1+\sin A}
\displaystyle =\frac{1-\sin A}{1+\sin A}\cdot\frac{1-\sin A}{1-\sin A}
\displaystyle =\frac{(1-\sin A)^2}{1-\sin^2A}
\displaystyle =\frac{1+\sin^2A-2\sin A}{\cos^2A}
\displaystyle =\frac{2-\cos^2A-2\sin A}{\cos^2A}
\displaystyle =\frac{2}{\cos^2A}-1-\frac{2\sin A}{\cos^2A}
\displaystyle =2\sec^2A-1-2\left(\frac{1}{\cos A}\right)\left(\frac{\sin A}{\cos A}\right)
\displaystyle =2\sec^2A-1-2\sec A\tan A=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 21: }(\sin A+\mathrm{cosec}A)^2+(\cos A+\sec A)^2=7+\tan^2A+\cot^2A.
\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\cdot\mathrm{cosec}A+\cos^2A+\sec^2A+2\cos A\cdot\sec A
\displaystyle =\sin^2A+\mathrm{cosec}^2A+2+\cos^2A+\sec^2A+2
\displaystyle =(\sin^2A+\cos^2A)+\mathrm{cosec}^2A+\sec^2A+4
\displaystyle =5+\mathrm{cosec}^2A+\sec^2A
\displaystyle =5+(1+\cot^2A)+(1+\tan^2A)
\displaystyle =7+\tan^2A+\cot^2A=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 22: }\sec^2A\cdot\mathrm{cosec}^2A=\tan^2A+\cot^2A+2.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sec^2A\cdot\mathrm{cosec}^2A
\displaystyle =(1+\tan^2A)(1+\cot^2A)
\displaystyle =1+\tan^2A+\cot^2A+\tan^2A\cot^2A
\displaystyle =1+\tan^2A+\cot^2A+1
\displaystyle =\tan^2A+\cot^2A+2=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 23: }\frac{1}{1+\cos A}+\frac{1}{1-\cos A}=2\mathrm{cosec}^2A.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{1}{1+\cos A}+\frac{1}{1-\cos A}
\displaystyle =\frac{1-\cos A+1+\cos A}{(1+\cos A)(1-\cos A)}
\displaystyle =\frac{2}{1-\cos^2A}
\displaystyle =\frac{2}{\sin^2A}
\displaystyle =2\mathrm{cosec}^2A=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 24: }\frac{1}{1-\sin A}+\frac{1}{1+\sin A}=2\sec^2A.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{1}{1-\sin A}+\frac{1}{1+\sin A}
\displaystyle =\frac{1+\sin A+1-\sin A}{(1-\sin A)(1+\sin A)}
\displaystyle =\frac{2}{1-\sin^2A}
\displaystyle =\frac{2}{\cos^2A}
\displaystyle =2\sec^2A=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 25: }\frac{\mathrm{cosec}A}{\mathrm{cosec}A-1}+\frac{\mathrm{cosec}A}{\mathrm{cosec}A+1}=2\sec^2A.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\mathrm{cosec}A}{\mathrm{cosec}A-1}+\frac{\mathrm{cosec}A}{\mathrm{cosec}A+1}
\displaystyle =\frac{\frac{1}{\sin A}}{\frac{1}{\sin A}-1}+\frac{\frac{1}{\sin A}}{\frac{1}{\sin A}+1}
\displaystyle =\frac{1}{1-\sin A}+\frac{1}{1+\sin A}
\displaystyle =\frac{1+\sin A+1-\sin A}{(1-\sin A)(1+\sin A)}
\displaystyle =\frac{2}{1-\sin^2A}
\displaystyle =\frac{2}{\cos^2A}
\displaystyle =2\sec^2A=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 26: }\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{\frac{1}{\cos A}}{\frac{1}{\cos A}-1}+\frac{\frac{1}{\cos A}}{\frac{1}{\cos A}+1}
\displaystyle =\frac{1}{1-\cos A}+\frac{1}{1+\cos A}
\displaystyle =\frac{1+\cos A+1-\cos A}{(1-\cos A)(1+\cos A)}
\displaystyle =\frac{2}{1-\cos^2A}
\displaystyle =\frac{2}{\sin^2A}
\displaystyle =2\mathrm{cosec}^2A=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 27: }\frac{1+\cos A}{1-\cos A}=\frac{\tan^2A}{(\sec A-1)^2}.
\displaystyle \text{Answer:}
\displaystyle \text{RHS}=\frac{\tan^2A}{(\sec A-1)^2}
\displaystyle =\frac{\frac{\sin^2A}{\cos^2A}}{\left(\frac{1}{\cos A}-1\right)^2}
\displaystyle =\frac{\sin^2A}{\cos^2A}\cdot\frac{\cos^2A}{(1-\cos A)^2}
\displaystyle =\frac{\sin^2A}{(1-\cos A)^2}
\displaystyle =\frac{(1-\cos A)(1+\cos A)}{(1-\cos A)^2}
\displaystyle =\frac{1+\cos A}{1-\cos A}=\text{LHS}.
\displaystyle \therefore \text{RHS}=\text{LHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 28: }\frac{1-\sin A}{1+\sin A}=\frac{\cot^2A}{(\mathrm{cosec}A+1)^2}.
\displaystyle \text{Answer:}
\displaystyle \text{RHS}=\frac{\cot^2A}{(\mathrm{cosec}A+1)^2}
\displaystyle =\frac{\frac{\cos^2A}{\sin^2A}}{\left(\frac{1}{\sin A}+1\right)^2}
\displaystyle =\frac{\cos^2A}{\sin^2A}\cdot\frac{\sin^2A}{(1+\sin A)^2}
\displaystyle =\frac{\cos^2A}{(1+\sin A)^2}
\displaystyle =\frac{(1-\sin A)(1+\sin A)}{(1+\sin A)^2}
\displaystyle =\frac{1-\sin A}{1+\sin A}=\text{LHS}.
\displaystyle \therefore \text{RHS}=\text{LHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 29: }\frac{1+\sin A}{\cos A}+\frac{\cos A}{1+\sin A}=2\sec A.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{1+\sin A}{\cos A}+\frac{\cos A}{1+\sin A}
\displaystyle =\frac{(1+\sin A)^2+\cos^2A}{\cos A(1+\sin A)}
\displaystyle =\frac{1+\sin^2A+2\sin A+\cos^2A}{\cos A(1+\sin A)}
\displaystyle =\frac{2+2\sin A}{\cos A(1+\sin A)}
\displaystyle =\frac{2(1+\sin A)}{\cos A(1+\sin A)}
\displaystyle =\frac{2}{\cos A}
\displaystyle =2\sec A=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 30: }\frac{1-\sin A}{1+\sin A}=(\sec A-\tan A)^2.
\displaystyle \text{Answer:}
\displaystyle \text{RHS}=(\sec A-\tan A)^2
\displaystyle =\left(\frac{1}{\cos A}-\frac{\sin A}{\cos A}\right)^2
\displaystyle =\frac{(1-\sin A)^2}{\cos^2A}
\displaystyle =\frac{(1-\sin A)^2}{(1-\sin A)(1+\sin A)}
\displaystyle =\frac{1-\sin A}{1+\sin A}=\text{LHS}.
\displaystyle \therefore \text{RHS}=\text{LHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 31: }\frac{1-\cos A}{1+\cos A}=(\cot A-\mathrm{cosec}A)^2.
\displaystyle \text{Answer:}
\displaystyle \text{RHS}=(\cot A-\mathrm{cosec}A)^2
\displaystyle =\left(\frac{\cos A}{\sin A}-\frac{1}{\sin A}\right)^2
\displaystyle =\frac{(\cos A-1)^2}{\sin^2A}
\displaystyle =\frac{(1-\cos A)^2}{(1-\cos A)(1+\cos A)}
\displaystyle =\frac{1-\cos A}{1+\cos A}=\text{LHS}.
\displaystyle \therefore \text{RHS}=\text{LHS. Hence proved.}
\displaystyle \

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

\displaystyle \textbf{Question 33: }\tan^2A-\tan^2B=\frac{\sin^2A-\sin^2B}{\cos^2A\cdot\cos^2B}.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\tan^2A-\tan^2B
\displaystyle =\frac{\sin^2A}{\cos^2A}-\frac{\sin^2B}{\cos^2B}
\displaystyle =\frac{\sin^2A\cos^2B-\sin^2B\cos^2A}{\cos^2A\cdot\cos^2B}
\displaystyle =\frac{\sin^2A(1-\sin^2B)-\sin^2B(1-\sin^2A)}{\cos^2A\cdot\cos^2B}
\displaystyle =\frac{\sin^2A-\sin^2A\sin^2B-\sin^2B+\sin^2A\sin^2B}{\cos^2A\cdot\cos^2B}
\displaystyle =\frac{\sin^2A-\sin^2B}{\cos^2A\cdot\cos^2B}=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 34: }\frac{\sin A-2\sin^3A}{2\cos^3A-\cos A}=\tan A.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\sin A-2\sin^3A}{2\cos^3A-\cos A}
\displaystyle =\frac{\sin A(1-2\sin^2A)}{\cos A(2\cos^2A-1)}
\displaystyle =\frac{\sin A}{\cos A}\cdot\frac{1-2\sin^2A}{2\cos^2A-1}
\displaystyle =\frac{\sin A}{\cos A}\cdot\frac{1-\sin^2A-\sin^2A}{\cos^2A+\cos^2A-1}
\displaystyle =\frac{\sin A}{\cos A}\cdot\frac{\cos^2A-\sin^2A}{\cos^2A-\sin^2A}
\displaystyle =\frac{\sin A}{\cos A}
\displaystyle =\tan A=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 35: }\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}\cdot\frac{1+\cos A}{1+\cos A}
\displaystyle =\frac{1-\cos^2A}{\sin A(1+\cos A)}
\displaystyle =\frac{\sin^2A}{\sin A(1+\cos A)}
\displaystyle =\frac{\sin A}{1+\cos A}=\text{LHS}.
\displaystyle \therefore \text{RHS}=\text{LHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 36: }\frac{\cos A}{1-\sin A}=\sec A+\tan A.
\displaystyle \text{Answer:}
\displaystyle \text{RHS}=\sec A+\tan A
\displaystyle =\frac{1}{\cos A}+\frac{\sin A}{\cos A}
\displaystyle =\frac{1+\sin A}{\cos A}
\displaystyle =\frac{1+\sin A}{\cos A}\cdot\frac{1-\sin A}{1-\sin A}
\displaystyle =\frac{1-\sin^2A}{\cos A(1-\sin A)}
\displaystyle =\frac{\cos^2A}{\cos A(1-\sin A)}
\displaystyle =\frac{\cos A}{1-\sin A}=\text{LHS}.
\displaystyle \therefore \text{RHS}=\text{LHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 37: }\frac{\sin A\cdot\tan A}{1-\cos A}=1+\sec A.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\sin A\cdot\tan A}{1-\cos A}
\displaystyle =\frac{\sin A}{1-\cos A}\cdot\frac{\sin A}{\cos A}
\displaystyle =\frac{\sin^2A}{(1-\cos A)\cos A}
\displaystyle =\frac{(1-\cos A)(1+\cos A)}{(1-\cos A)\cos A}
\displaystyle =\frac{1+\cos A}{\cos A}
\displaystyle =\frac{1}{\cos A}+1
\displaystyle =1+\sec A=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 38: }(1+\cot A-\mathrm{cosec}A)(1+\tan A+\sec A)=2.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(1+\cot A-\mathrm{cosec}A)(1+\tan A+\sec A)
\displaystyle =\left(1+\frac{\cos A}{\sin A}-\frac{1}{\sin A}\right)\left(1+\frac{\sin A}{\cos A}+\frac{1}{\cos A}\right)
\displaystyle =\frac{\sin A+\cos A-1}{\sin A}\cdot\frac{\sin A+\cos A+1}{\cos A}
\displaystyle =\frac{(\sin A+\cos A)^2-1}{\sin A\cos A}
\displaystyle =\frac{\sin^2A+\cos^2A+2\sin A\cos A-1}{\sin A\cos A}
\displaystyle =\frac{1+2\sin A\cos A-1}{\sin A\cos A}
\displaystyle =\frac{2\sin A\cos A}{\sin A\cos A}
\displaystyle =2=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 39: }\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sqrt{\frac{1+\sin A}{1-\sin A}}
\displaystyle =\sqrt{\frac{1+\sin A}{1-\sin A}\cdot\frac{1+\sin A}{1+\sin A}}
\displaystyle =\sqrt{\frac{(1+\sin A)^2}{1-\sin^2A}}
\displaystyle =\sqrt{\frac{(1+\sin A)^2}{\cos^2A}}
\displaystyle =\frac{1+\sin A}{\cos A}\quad[\text{Since }A\text{ is acute}]
\displaystyle =\frac{1}{\cos A}+\frac{\sin A}{\cos A}
\displaystyle =\sec A+\tan A=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 40: }\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}\cdot\frac{1-\cos A}{1-\cos A}}
\displaystyle =\sqrt{\frac{(1-\cos A)^2}{1-\cos^2A}}
\displaystyle =\sqrt{\frac{(1-\cos A)^2}{\sin^2A}}
\displaystyle =\frac{1-\cos A}{\sin A}\quad[\text{Since }A\text{ is acute}]
\displaystyle =\frac{1}{\sin A}-\frac{\cos A}{\sin A}
\displaystyle =\mathrm{cosec}A-\cot A=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 41: }\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}\cdot\frac{1+\cos A}{1+\cos A}}
\displaystyle =\sqrt{\frac{1-\cos^2A}{(1+\cos A)^2}}
\displaystyle =\sqrt{\frac{\sin^2A}{(1+\cos A)^2}}
\displaystyle =\frac{\sin A}{1+\cos A}\quad[\text{Since }A\text{ is acute}]
\displaystyle =\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 42: }\sqrt{\frac{1-\sin A}{1+\sin A}}=\frac{\cos A}{1+\sin A}.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sqrt{\frac{1-\sin A}{1+\sin A}}
\displaystyle =\sqrt{\frac{1-\sin A}{1+\sin A}\cdot\frac{1+\sin A}{1+\sin A}}
\displaystyle =\sqrt{\frac{1-\sin^2A}{(1+\sin A)^2}}
\displaystyle =\sqrt{\frac{\cos^2A}{(1+\sin A)^2}}
\displaystyle =\frac{\cos A}{1+\sin A}\quad[\text{Since }A\text{ is acute}]
\displaystyle =\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 43: }1-\frac{\cos^2A}{1+\sin A}=\sin A\hspace{1.0cm}[\text{ICSE }2001]
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=1-\frac{\cos^2A}{1+\sin A}
\displaystyle =\frac{1+\sin A-\cos^2A}{1+\sin A}
\displaystyle =\frac{1+\sin A-(1-\sin^2A)}{1+\sin A}
\displaystyle =\frac{\sin A+\sin^2A}{1+\sin A}
\displaystyle =\frac{\sin A(1+\sin A)}{1+\sin A}
\displaystyle =\sin A=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 44: }\frac{1}{\sin A+\cos A}+\frac{1}{\sin A-\cos A}=\frac{2\sin A}{1-2\cos^2A}\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 A+\cos A)(\sin A-\cos A)}
\displaystyle =\frac{2\sin A}{\sin^2A-\cos^2A}
\displaystyle =\frac{2\sin A}{1-\cos^2A-\cos^2A}
\displaystyle =\frac{2\sin A}{1-2\cos^2A}=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 45: }\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{2}{2\sin^2A-1}.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}
\displaystyle =\frac{(\sin A+\cos A)^2+(\sin A-\cos A)^2}{(\sin A-\cos A)(\sin A+\cos A)}
\displaystyle =\frac{\sin^2A+\cos^2A+2\sin A\cos A+\sin^2A+\cos^2A-2\sin A\cos A}{\sin^2A-\cos^2A}
\displaystyle =\frac{2(\sin^2A+\cos^2A)}{\sin^2A-\cos^2A}
\displaystyle =\frac{2}{\sin^2A-\cos^2A}
\displaystyle =\frac{2}{\sin^2A-(1-\sin^2A)}
\displaystyle =\frac{2}{2\sin^2A-1}=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 46: }\frac{\cot A+\mathrm{cosec}A-1}{\cot A-\mathrm{cosec}A+1}=\frac{1+\cos A}{\sin A}.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\cot A+\mathrm{cosec}A-1}{\cot A-\mathrm{cosec}A+1}
\displaystyle =\frac{\frac{\cos A}{\sin A}+\frac{1}{\sin A}-1}{\frac{\cos A}{\sin A}-\frac{1}{\sin A}+1}
\displaystyle =\frac{\cos A+1-\sin A}{\cos A-1+\sin A}
\displaystyle =\frac{\cos A+1-\sin A}{\cos A-1+\sin A}\cdot\frac{\cos A+1+\sin A}{\cos A+1+\sin A}
\displaystyle =\frac{(\cos A+1)^2-\sin^2A}{(\cos A+\sin A)^2-1}
\displaystyle =\frac{\cos^2A+1+2\cos A-\sin^2A}{\cos^2A+\sin^2A+2\sin A\cos A-1}
\displaystyle =\frac{2\cos^2A+2\cos A}{2\sin A\cos A}
\displaystyle =\frac{2\cos A(1+\cos A)}{2\sin A\cos A}
\displaystyle =\frac{1+\cos A}{\sin A}=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 47: }\frac{\sin\theta\cdot\tan\theta}{1-\cos\theta}=1+\sec\theta\hspace{1.0cm}[\text{ICSE }2006]
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\sin\theta\cdot\tan\theta}{1-\cos\theta}
\displaystyle =\frac{\sin\theta}{1-\cos\theta}\cdot\frac{\sin\theta}{\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 =\frac{1}{\cos\theta}+1
\displaystyle =\sec\theta+1=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

\displaystyle \textbf{Question 48: }\frac{\cos\theta\cdot\cot\theta}{1+\sin\theta}=\mathrm{cosec}\theta-1.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\cos\theta\cdot\cot\theta}{1+\sin\theta}
\displaystyle =\frac{\cos\theta}{1+\sin\theta}\cdot\frac{\cos\theta}{\sin\theta}
\displaystyle =\frac{\cos^2\theta}{\sin\theta(1+\sin\theta)}
\displaystyle =\frac{(1-\sin\theta)(1+\sin\theta)}{\sin\theta(1+\sin\theta)}
\displaystyle =\frac{1-\sin\theta}{\sin\theta}
\displaystyle =\frac{1}{\sin\theta}-1
\displaystyle =\mathrm{cosec}\theta-1=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \

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