Class 10: Trigonometry – Exercise 23(a) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Prove the following identities:

$\displaystyle \text{Question 1: } \frac{\sec A -1 }{\sec A + 1} = \frac{1- \cos A }{1+ \cos A} \hspace{1.0cm} [2007]$

Answer:

$\displaystyle \text{LHS } = \frac{\sec A -1 }{\sec A + 1} = \frac{\frac{1}{\cos A} -1 }{\frac{1}{\cos A} + 1} = \frac{1- \cos A }{1+ \cos A} = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 2: } \frac{1+ \sin A }{1- \sin A} = \frac{\mathrm{cosec} A + 1 }{\mathrm{cosec} A - 1}$

Answer:

$\displaystyle \text{RHS } = \frac{ \mathrm{cosec} A + 1 }{\mathrm{cosec} A - 1} = \frac{\frac{1}{\sin A} +1 }{\frac{1}{\sin A} - 1} = \frac{1+\sin A }{1- \sin A} = \text{ LHS. Hence Proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 3: } \frac{1}{\tan A + \cot A} = \cos A \sin A$

Answer:

$\displaystyle \text{LHS } = \frac{1}{\tan A + \cot A} = \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} = \frac{\cos A . \sin A}{\sin^2 A+ \cos^2 A} \\ \\ = \cos A . \sin A = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 4: } \tan A - \cot A = \frac{1- 2 \cos^2 A}{\sin A . \cos A}$

Answer:

$\displaystyle \text{LHS } = \tan A - \cot A$

$\displaystyle = \frac{\sin A}{\cos A} - \frac{\cos A}{\sin A} = \frac{\sin^2 A - \cos^2 A}{\sin A . \cos A} = \frac{1 - \cos^2 A - \cos^2 A}{\sin A . \cos A} = \frac{1- 2 \cos^2 A}{\sin A . \cos A}$

$\displaystyle = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 5: } \sin^4 A - \cos^4 A = 2 \sin^2 A - 1$

Answer:

$\displaystyle \text{LHS } = \sin^4 A - \cos^4 A$

$\displaystyle = (\sin^2 A - \cos^2 A)(\sin^2 A + \cos^2 A)$

$\displaystyle = \sin^2 A - \cos^2 A$

$\displaystyle = 1- \cos^2 A - \cos^2 A = 1- 2 \cos^2 A = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 6: } (1-\tan A)^2 + (1 - \tan A)^2 = 2 \sec^2 A \hspace{1.0cm} [2005]$

Answer:

$\displaystyle \text{LHS } = (1-\tan A)^2 + (1 - \tan A)^2$

$\displaystyle = (1- \frac{\sin A}{\cos A} )^2 + (1+ \frac{\sin A}{\cos A} )^2$

$\displaystyle = \frac{(\cos A - \sin A)^2}{\cos^2 A} + \frac{(\cos A + \sin A)^2}{\cos^2 A}$

$\displaystyle = \frac{\cos^2 A + \sin^2 A - 2 \cos A . \sin A+ \cos^2 A + \sin^2 A + 2 \cos A . \sin A}{\cos^2 A}$

$\displaystyle = \frac{2}{\cos^2 A} = 2 \sec^2 A = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 7: } \mathrm{cosec}^4 A - \mathrm{cosec}^2 A = \cot^4 A + \cot^2 A$

Answer:

$\displaystyle \text{LHS } = \mathrm{cosec}^4 A - \mathrm{cosec}^2 A$

$\displaystyle = \frac{\cos^4 A}{\sin^4 A} + \frac{\cos^2 A}{\sin ^2 A}$

$\displaystyle = \frac{\cos^2 A}{\sin^2 A} \Big( \frac{\cos^2 A}{\sin^2 A} + 1\Big)$

$\displaystyle = \frac{\cos^2 A}{\sin^4 A}$

$\displaystyle = \frac{1 - \sin^2 A}{\sin^4 A}$

$\displaystyle = \frac{1}{\sin^4 A} - \frac{1}{\sin^2 A}$

$\displaystyle = \mathrm{cosec}^4 A - \mathrm{cosec}^2 A = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 8: } \sec A(1- \sin A) (\sec A + \tan A) = 1$

Answer:

$\displaystyle \text{LHS } = \sec A(1- \sin A) (\sec A + \tan A)$

$\displaystyle = \frac{1}{\cos A} (1- \sin A) \Bigg( \frac{1}{\cos A} + \frac{\sin A}{\cos A} \Bigg)$

$\displaystyle = \frac{1 - \sin^2 A}{\cos^2 A}$

$\displaystyle = \frac{\cos^2 A}{\cos^2 A} = 1 = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 9: } \mathrm{cosec} A (1+ \cos A) (\mathrm{cosec} A - \cot A) = 2$

Answer:

$\displaystyle \text{LHS } = \mathrm{cosec} A (1+ \cos A) (\mathrm{cosec} A - \cot A)$

$\displaystyle = \frac{1}{\sin A} (1 + \cos A) (\frac{1}{\sin A} - \frac{\cos A}{\sin A} )$

$\displaystyle = \frac{1}{\sin^2 A} (1+ \cos A)(1- \cos A)$

$\displaystyle = \frac{1-\cos^2 A}{\sin^2 A}$

$\displaystyle = \frac{\sin^2 A}{\sin^2 A} = 1 = \text{ RHS }$

$\displaystyle \\$

$\displaystyle \text{Question 10: } \sec^2 A + \mathrm{cosec}^2 A = \sec^2 A. \mathrm{cosec}^2 A$

Answer:

$\displaystyle \text{LHS } = \sec^2 A + \mathrm{cosec}^2 A$

$\displaystyle = \frac{1}{\cos^2 A} + \frac{1}{\sin^2 A}$

$\displaystyle = \frac{\sin^2 A + \cos^2 A}{\sin^2 A. \cos^2 A}$

$\displaystyle = \sec^2 A . \mathrm{cosec}^2 A = \text{ RHS Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 11: } \frac{(1+\tan^2 A) \cot A}{ \mathrm{cosec}^2 A} = \tan A$

Answer:

$\displaystyle \text{LHS } = \frac{(1+\tan^2 A) \cot A}{ \mathrm{cosec}^2 A}$

$\displaystyle = \frac{\cos^2 A + \sin^2 A}{\cos^2 A} . \frac{\cos A}{\sin A} . \sin^2 A$

$\displaystyle = \frac{1}{\cos A} \sin A$

$\displaystyle = \tan A = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 12: } \tan^2 A - \sin^2 A= \tan^2 A. \sin^2 A$

Answer:

$\displaystyle \text{LHS } = \tan^2 A - \sin^2 A$

$\displaystyle = \frac{\sin^2 A}{\cos^2 A} - \sin^2 A$

$\displaystyle = \frac{\sin^2 A(1-\cos^2 A)}{\cos^2 A}$

$\displaystyle = \frac{\sin^4 A}{\cos^2 A} = \tan^2 A. \sin^2 A= \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 13: } \cot^2 A - \cos^2 A = \cos^2 A .\cot^2 A$

Answer:

$\displaystyle \text{LHS } = \cot^2 A - \cos^2 A$

$\displaystyle = \frac{\cos^2 A}{\sin^2 A} - \cos^2 A$

$\displaystyle = \frac{\cos^2 A(1-\sin^2 A)}{\sin^2 A}$

$\displaystyle = \frac{\cos^4 A}{\sin^2 A} = \cot^2 A. \cos^2 A= \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 14: } (\mathrm{cosec} A + \sin A)(\mathrm{cosec} A - \sin A) = \cot^2 A + \cos^ A$

Answer:

$\displaystyle \text{LHS } = (\mathrm{cosec} A + \sin A)(\mathrm{cosec} A - \sin A)$

$\displaystyle = \Big( \frac{1}{\sin A} + \sin A \Big) \Big( \frac{1}{\sin A} - \sin A \Big)$

$\displaystyle = \frac{1+\sin^2 A}{\sin A} . \frac{1 - \sin^2 A}{\sin A}$

$\displaystyle = \frac{\cos^2 A}{\sin^2 A} (1+ \sin^2 A)$

$\displaystyle = \cot^2 A + \cos^2 A= \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 15: } (\sec A - \cos A)(\sec A + \cos A)= \sin^2 A + \tan^2 A$

Answer:

$\displaystyle \text{LHS } = (\sec A - \cos A)(\sec A + \cos A)$

$\displaystyle = \Big( \frac{1}{\cos A} + \cos A \Big) \Big( \frac{1}{\cos A} - \cos A \Big)$

$\displaystyle = \frac{1+\cos^2 A}{\cos A} . \frac{1 - \cos^2 A}{\cos A}$

$\displaystyle = \frac{\sin^2 A}{\cos^2 A} (1 + \cos^2 A)$

$\displaystyle = \tan^2 A + \sin^2 A = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 16: } (\cos A + \sin A)^2 + (\cos A - \sin A)^2 = 2$

Answer:

$\displaystyle \text{LHS } = (\cos A + \sin A)^2 + (\cos A - \sin A)^2$

$\displaystyle = \cos^2 A + \sin^2 A + 2 \cos A .\sin A + \cos^2 A + \sin^2 A - 2 \cos A .\sin A$

$\displaystyle = 2(\cos^2 A + \sin^2 A) = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 17: } (\mathrm{cosec} A - \sin A) (\sec A - \cos A) (\tan A + \cot A) = 2$

Answer:

$\displaystyle \text{LHS } = (\mathrm{cosec} A - \sin A) (\sec A - \cos A) (\tan A + \cot A)$

$\displaystyle = \Big( \frac{1}{\sin A} - \sin A \Big) \Big( \frac{1}{\cos A} - \cos A \Big) \Big(\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} \Big)$

$\displaystyle = \frac{1-\sin^2 A}{\sin A} . \frac{1-\cos^2 A}{\cos A} . \frac{1}{\sin A . \cos A}$

$\displaystyle = \frac{\cos^2 A}{\sin A} . \frac{\sin^2 A}{\cos A} . \frac{1}{\sin A . \cos A}$

$\displaystyle = 1 = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 18: } \frac{1}{\sec A + \tan A} = \sec A - \tan A$

Answer:

$\displaystyle \text{LHS } = \frac{1}{\sec A + \tan A}$

$\displaystyle = \frac{1}{\frac{1}{\cos A}+\frac{\sin A}{\cos A}}$

$\displaystyle = \frac{\cos A}{1+\sin A}$

$\displaystyle = \frac{\cos A}{1+\sin A} \times \frac{1- \sin A}{1- \sin A}$

$\displaystyle = \frac{\cos A(1- \sin A)}{\cos^2 A}$

$\displaystyle = \frac{1-\sin A}{\cos A} = \sec A - \tan A = \text{ RHS }$

$\displaystyle \\$

$\displaystyle \text{Question 19: } \mathrm{cosec} A + \cot A = \frac{1}{\mathrm{cosec} A - \cot A}$

Answer:

$\displaystyle \text{LHS } = \mathrm{cosec} A + \cot A$

$\displaystyle = \frac{1}{\sin A} + \frac{\cos A}{\sin A}$

$\displaystyle = \frac{1+\cos A}{\sin A}$

$\displaystyle = \frac{1+\cos A}{\sin A} \times \frac{1-\cos A}{1-\cos A}$

$\displaystyle = \frac{\sin A}{1 - \cos A}$

$\displaystyle \text{RHS } = \frac{1}{\frac{1}{\sin A} - \frac{\cos A}{\sin A}}$

$\displaystyle = \frac{\sin A}{1 - \cos A}$

Therefore LHS = RHS. Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 20: } \frac{\sec A - \tan A}{\sec A + \tan A} = 2 \sec^2 A - 1-2 \sec A \tan A$

Answer:

$\displaystyle \text{LHS } = \frac{\sec A - \tan A}{\sec A + \tan A}$

$\displaystyle = \frac{\frac{1}{\cos A} - \frac{\sin A}{\cos A}}{\frac{1}{\cos A} + \frac{\sin A}{\cos A}}$

$\displaystyle = \frac{1-\sin A}{1+ \sin A}$

$\displaystyle = \frac{1-\sin A}{1+ \sin A} \times \frac{1-\sin A}{1- \sin A}$

$\displaystyle = \frac{1 + \sin^2 A - 2 \sin A}{1 - \sin^2 A}$

$\displaystyle = \frac{2 - \cos^2 - 2 \sin A}{\cos^2 A}$

$\displaystyle = 2 \sec^2 A - 1-2 \sec A . \tan A$

$\displaystyle \\$

$\displaystyle \text{Question 21: } (\sin A + \mathrm{cosec} A)^2 +(\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$

Answer:

$\displaystyle \text{LHS } = (\sin A + \mathrm{cosec} A)^2 +(\cos A + \sec A)^2$

$\displaystyle = \sin^2 A + \mathrm{cosec}^2 A + 2 + \cos^2 A + \sec^2 A + 2$

$\displaystyle = 5 + \mathrm{cosec}^2 A + \sec^2 A$

$\displaystyle = 5 + (1 + \cot^2 A) + (1 + \tan^2 A)$

$\displaystyle = 7 + \tan^2 A + \cot^2 A = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 22: } \sec^2 A . \mathrm{cosec}^2 A = \tan^2 A + \cot^2 A + 2$

Answer:

$\displaystyle \text{LHS } = \sec^2 A . \mathrm{cosec}^2 A$

$\displaystyle = (1 +\tan^2 A)(1 + \cot^2 A)$

$\displaystyle = 1 +\tan^2 A+ \cot^2 A + 1$

$\displaystyle = \tan^2 A + \cot^2 A + 2 = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 23: } \frac{1}{1+\cos A} + \frac{1}{1- \cos A} = 2 \mathrm{cosec}^2 A$

Answer:

$\displaystyle \text{LHS } = \frac{1}{1+\cos A} + \frac{1}{1- \cos A}$

$\displaystyle = \frac{1- \cos A + 1 + \cos A}{1- \cos^2 A}$

$\displaystyle = \frac{2}{\sin^2 A} = 2 \mathrm{cosec}^2 A = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 24: } \frac{1}{1 - \sin A} + \frac{1}{1+\sin A} = 2 \sec^2 A$

Answer:

$\displaystyle \text{LHS } = \frac{1}{1 - \sin A} + \frac{1}{1+\sin A}$

$\displaystyle = \frac{1- \sin A + 1 + \sin A}{1- \sin^2 A}$

$\displaystyle = \frac{2}{\cos^2 A} = 2 \sec^2 A = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 25: } \frac{\mathrm{cosec} A}{\mathrm{cosec} A - 1} + \frac{\mathrm{cosec} A}{\mathrm{cosec} A + 1} = 2 \sec^2 A$

Answer:

$\displaystyle \text{LHS } = \frac{\mathrm{cosec} A}{\mathrm{cosec} A - 1} + \frac{\mathrm{cosec} A}{\mathrm{cosec} A + 1}$

$\displaystyle = \frac{1}{1- \sin A} + \frac{1}{1+ \sin A}$

$\displaystyle = \frac{1+ \sin A + 1- \sin A }{1- \sin^2 A}$

$\displaystyle = \frac{2}{\cos^2 A}$

$\displaystyle = 2 \sec^2 A = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 26: } \frac{\sec A}{\sec A - 1} + \frac{\sec A}{\sec A + 1} = 2 \mathrm{cosec}^2 A$

Answer:

$\displaystyle \text{LHS } = \frac{\sec A}{\sec A - 1} + \frac{\sec A}{\sec A + 1}$

$\displaystyle = \frac{1}{1- \cos A} + \frac{1}{1+ \cos A}$

$\displaystyle = \frac{1+ \cos A + 1- \cos A }{1- \cos^2 A}$

$\displaystyle = \frac{2}{\sin^2 A}$

$\displaystyle = 2 \mathrm{cosec}^2 A = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 27: } \frac{1+ \cos A}{1- \cos A} = \frac{\tan^2 A}{(\sec A - 1)^2}$

Answer:

$\displaystyle \text{RHS } = \frac{\tan^2 A}{(\sec A - 1)^2}$

$\displaystyle = \frac{\sin^2 A}{\cos^2 A} \times \frac{\cos^2 A}{(1-\cos A)^2}$

$\displaystyle = \frac{(1-\cos A)(1 + \cos A)}{(1-\cos A)^2}$

$\displaystyle = \frac{1+ \cos A}{1- \cos A} = \text{ LHS. Hence proved.}$

$\displaystyle \\$

$\displaystyle \text{Question 28: } \frac{1- \sin A}{1 + \sin A} = \frac{\cot^2 A}{(\mathrm{cosec} A - 1)^2}$

Answer:

$\displaystyle \text{RHS } = \frac{\cot^2 A}{(\mathrm{cosec} A - 1)^2}$

$\displaystyle = \frac{\cos^2 A}{\sin^2 A} \times \frac{\sin^2 A}{(1-\sin A)^2}$

$\displaystyle = \frac{(1-\sin A)(1 + \sin A)}{(1+\sin A)^2}$

$\displaystyle = \frac{1- \sin A}{1+ \sin A} = \text{ LHS. Hence proved.}$

$\displaystyle \\$

$\displaystyle \text{Question 29: } \frac{1 + \sin A}{\cos A} + \frac{\cos A}{1 + \sin A} = 2 \sec A$

Answer:

$\displaystyle \text{LHS } = \frac{1 + \sin A}{\cos A} + \frac{\cos A}{1 + \sin A}$

$\displaystyle = \frac{1 + \sin^2 A + 2 \sin A + \cos^2 A}{\cos A (1 + \sin A)}$

$\displaystyle = \frac{2 + 2 \sin A}{\cos A (1 + \sin A)}$

$\displaystyle = \frac{2}{\cos A}$

$\displaystyle = 2 \sec A = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 30: } \frac{1- \sin A}{1 + \sin A} = (\sec A - \tan A)^2$

Answer:

$\displaystyle \text{RHS } = (\sec A - \tan A)^2$

$\displaystyle = \Big( \frac{1}{\cos A} - \frac{\sin A}{\cos A} \Big)^2$

$\displaystyle = \frac{(1- \sin A)^2}{\cos^2 A}$

$\displaystyle = \frac{1- \sin A}{1+ \sin A} = \text{ LHS. Hence proved.}$

$\displaystyle \\$

$\displaystyle \text{Question 31: } \frac{1- \cos A}{1 + \cos A} = (\cot A - \mathrm{cosec} A)^2$

Answer:

$\displaystyle \text{RHS } = (\cot A - \mathrm{cosec} A)^2$

$\displaystyle = \Big( \frac{\cos A}{\sin A} - \frac{1}{\sin A} \Big)^2$

$\displaystyle = \frac{( \cos A - 1)^2}{\sin^2 A}$

$\displaystyle = \frac{1- \cos A}{1+ \cos A} = \text{ LHS. Hence proved.}$

$\displaystyle \\$

$\displaystyle \text{Question 32: } \frac{\mathrm{cosec} A - 1}{\mathrm{cosec} A + 1} = ( \frac{\cos A}{1 + \sin A} )^2$

Answer:

$\displaystyle \text{LHS } = \frac{\mathrm{cosec} A - 1}{\mathrm{cosec} A + 1}$

$\displaystyle = \frac{1 - \sin A}{1 + \sin A}$

$\displaystyle = \frac{1 - \sin A}{1 + \sin A} \times \frac{1 + \sin A}{1 + \sin A}$

$\displaystyle = \frac{1 - \sin^2 A}{(1 + \sin A)^2}$

$\displaystyle = \Big( \frac{\cos A}{1+ \sin A} \Big)^2$

$\displaystyle \\$

$\displaystyle \text{Question 33: } \tan^2 A - \tan^2 B = \frac{\sin^2 A - \sin^2 B}{\cos^2 A. \cos^2 B}$

Answer:

$\displaystyle \text{LHS } = \tan^2 A - \tan^2 B$

$\displaystyle = \frac{\sin^2 A}{\cos^2 A} - \frac{\sin^2 B}{\cos^2 B}$

$\displaystyle = \frac{\sin^2 A. \cos^2 B -\sin^2 B . \cos^2 A}{\cos^2 A . \cos^2 B}$

$\displaystyle = \frac{\sin^2 A. (1 - \sin^2 B) -\sin^2 B . (1 - \sin^2 A)}{\cos^2 A . \cos^2 B}$

$\displaystyle = \frac{\sin^2 A - \sin^2 B}{\cos^2 A. \cos^2 B} = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 34: } \frac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A} = \tan A$

Answer:

$\displaystyle \text{LHS } = \frac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A}$

$\displaystyle = \frac{\sin A( 1 - 2 \sin^2 A)}{\cos A(2 \cos^2 A - 1)}$

$\displaystyle = \frac{\sin A}{\cos A} . \frac{1 - 2 \sin^2 A}{2 \cos^2 A - 1}$

$\displaystyle = \frac{\sin A}{\cos A} . \frac{1 - \sin^2 A - \sin^2 A}{\cos^2 A + \cos^2 A -1}$

$\displaystyle = \frac{\sin A}{\cos A} . \frac{\cos^2 A - \sin^2 A}{\cos^2 A - \sin^2 A}$

$\displaystyle = \tan A = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 35: } \frac{\sin A}{1 + \cos A} = \mathrm{cosec} A - \cot A \hspace{1.0cm} [2008]$

Answer:

$\displaystyle \text{RHS } = \mathrm{cosec} A - \cot A$

$\displaystyle = \frac{1}{\sin A} - \frac{\cos A}{\sin A}$

$\displaystyle = \frac{1- \cos A}{\sin A}$

$\displaystyle = \frac{1- \cos A}{\sin A} \times \frac{1+ \cos A}{1+ \cos A}$

$\displaystyle = \frac{1 - \cos^2 A}{\sin A (1 + \cos A)}$

$\displaystyle = \frac{\sin A}{1 + \cos A} = \text{ LHS. Hence proved.}$

$\displaystyle \\$

$\displaystyle \text{Question 36: } \frac{\cos A}{1 - \sin A} = \sec A + \tan A$

Answer:

$\displaystyle \text{RHS } = \sec A + \tan A$

$\displaystyle = \frac{1}{\cos A} + \frac{\sin A}{\cos A}$

$\displaystyle = \frac{1+ \sin A}{\cos A}$

$\displaystyle = \frac{1+ \sin A}{\cos A} \times \frac{1- \sin A}{1- \sin A}$

$\displaystyle = \frac{(1 - \sin^2 A)}{\cos A (1 - \sin A)}$

$\displaystyle = \frac{\cos A}{1 - \sin A} = \text{ LHS. Hence proved.}$

$\displaystyle \\$

$\displaystyle \text{Question 37: } \frac{\sin A. \tan A}{1- \cos A} = 1+ \sec A$

Answer:

$\displaystyle \text{LHS } = \frac{\sin A. \tan A}{1- \cos A}$

$\displaystyle = \frac{\sin A}{1- \cos A} . \frac{\sin A}{\cos A}$

$\displaystyle = \frac{\sin^2 A}{(1 - \cos A) \cos A}$

$\displaystyle = \frac{(1- \cos A)(1 - \cos A)}{(1- \cos A) \cos A}$

$\displaystyle = \frac{1 + \cos A}{\cos A}$

$\displaystyle = 1 + \sec A$

$\displaystyle \\$

$\displaystyle \text{Question 38: } (1+\cot A - \mathrm{cosec} A)(1+\tan A+\sec A)=2$

Answer:

$\displaystyle \text{LHS } = (1+\cot A - \mathrm{cosec} A)(1+\tan A+\sec A)$

$\displaystyle = \Big(1+ \frac{\cos A}{\sin A} - \frac{1}{\sin A} \Big) \Big(1+ \frac{\sin A}{\cos A} + \frac{1}{\cos A} \Big)$

$\displaystyle = \frac{\sin A + \cos A - 1}{\sin A} \times \frac{\sin A + \cos A +1}{\cos A}$

$\displaystyle = \frac{(\sin A + \cos A)^2 -1}{\sin A . \cos A}$

$\displaystyle = \frac{\sin ^2 A + \cos^2 A + 2. \sin A. \cos A - 1}{\sin A . \cos A}$

$\displaystyle = \frac{2. \sin A. \cos A}{\sin A . \cos A}$

$\displaystyle = 2 = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 39: } \sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$

Answer:

$\displaystyle \text{LHS } = \sqrt{\frac{1 + \sin A}{1 - \sin A}}$

$\displaystyle = \sqrt{\frac{1 + \sin A}{1 - \sin A} \times \frac{1 + \sin A}{1 + \sin A}}$

$\displaystyle = \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}}$

$\displaystyle = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}}$

$\displaystyle = \frac{1+ \sin A}{\cos A}$

$\displaystyle = \sec A + \tan A = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 40: } \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \mathrm{cosec} A - \cot A \hspace{1.0cm} [2000]$

Answer:

$\displaystyle \text{LHS } = \sqrt{\frac{1 - \cos A}{1 + \cos A}}$

$\displaystyle = \sqrt{\frac{1 - \cos A}{1 + \cos A} \times \frac{1 - \cos A}{1 - \cos A}}$

$\displaystyle = \sqrt{\frac{(1 - \cos A)^2}{1 - \cos^2 A}}$

$\displaystyle = \sqrt{\frac{(1 - \cos A)^2}{\sin^2 A}}$

$\displaystyle = \frac{1- \cos A}{\sin A}$

$\displaystyle = \mathrm{cosec} A - \cot A = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 41: } \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\sin A}{1 + \cos A} \hspace{1.0cm} [2013]$

Answer:

$\displaystyle \text{LHS } = \sqrt{\frac{1 - \cos A}{1 + \cos A}}$

$\displaystyle = \sqrt{\frac{1 - \cos A}{1 + \cos A} \times \frac{1 + \cos A}{1 + \cos A}}$

$\displaystyle = \sqrt{\frac{1 - \cos^2 A}{(1 + \cos A)^2}}$

$\displaystyle = \frac{\sin A}{1 + \cos A} \text{ = RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 42: } \sqrt{\frac{1 - \sin A}{1 + \sin A}} = \frac{\cos A}{1 + \sin A}$

Answer:

$\displaystyle \text{LHS } = \sqrt{\frac{1 - \sin A}{1 + \sin A}}$

$\displaystyle = \sqrt{\frac{1 - \sin A}{1 + \sin A} \times \frac{1 + \sin A}{1 + \sin A}}$

$\displaystyle = \sqrt{\frac{1 - \sin^2 A}{(1 + \sin A)^2}}$

$\displaystyle = \frac{\cos A}{1 + \sin A} \text{ = RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 43: } 1 - \frac{\cos^2 A}{1 + \sin A} = \sin A \hspace{1.0cm} [2001]$

Answer:

$\displaystyle \text{LHS } = \frac{\cos^2 A}{1 + \sin A}$

$\displaystyle = \frac{1 + \sin A - \cos^2 A}{1 + \sin A}$

$\displaystyle = \frac{\sin A + \sin^2 A}{1 + \sin A}$

$\displaystyle = \frac{\sin A(1 + \sin A)}{1 + \sin A}$

$\displaystyle = \sin A = \text{RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 44: } \frac{1}{\sin A + \cos A} + \frac{1}{\sin A + \cos A} = \frac{2 \sin A}{1 - 2 \cos^2 A} \hspace{1.0cm} [2002]$

Answer:

$\displaystyle \text{LHS } = \frac{1}{\sin A + \cos A} + \frac{1}{\sin A + \cos A}$

$\displaystyle = \frac{\sin A - \cos A + \sin A + \cos A}{\sin^2 A - \cos^2 A}$

$\displaystyle = \frac{2 \sin A}{1 - 2 \cos^2 A} \text{ = RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 45: } \frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{2 \sin^2 A - 1}$

Answer:

$\displaystyle \text{LHS } = \frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A}$

$\displaystyle = \frac{(\sin A + \cos A)^2+(\sin A - \cos A)^2}{\sin^2 A - \cos^2 A}$

$\displaystyle = \frac{\sin^2 A + \cos^2 A + 2. \sin A . \cos A + \sin^2 A + \cos^2 A - 2. \sin A . \cos A }{1 - \cos^2 A - \cos^2 A}$

$\displaystyle = \frac{2}{1 - 2. \cos^2 A} \text{ = RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 46: } \frac{\cot A + \mathrm{cosec} A - 1}{\cot A - \mathrm{cosec} A + 1} = \frac{1 + \cos A}{\sin A}$

Answer:

$\displaystyle \text{LHS } = \frac{\cot A + \mathrm{cosec} A - 1}{\cot A - \mathrm{cosec} A + 1}$

$\displaystyle = \frac{\cos A + 1-\sin A}{\cos A - 1 + \sin A}$

$\displaystyle = \frac{\cos A + 1-\sin A}{\cos A - 1 + \sin A} \times \frac{\cos A+ 1 + \sin A}{\cos A + 1 + \sin A}$

$\displaystyle = \frac{\cos^2 A + 1 + 2 \cos A - \sin^2 A}{2 \cos A. \sin A}$

$\displaystyle = \frac{2 \cos^2 A + 2 \cos A}{2 \cos A. \sin A}$

$\displaystyle = \frac{1 + \cos A}{\sin A} \text{ = RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 47: } \frac{\sin \theta . \tan \theta}{1 - \cos \theta} = 1 + \sec \theta \hspace{1.0cm} [2006]$

Answer:

$\displaystyle \text{LHS } = \frac{\sin \theta . \tan \theta}{1 - \cos \theta}$

$\displaystyle = \frac{\sin^2 \theta}{\cos \theta (1 - \cos \theta)}$

$\displaystyle = \frac{(1-\cos \theta)(1 + \cos \theta)}{\cos \theta (1 - \cos \theta)}$

$\displaystyle = \frac{1 + \cos \theta}{\cos \theta}$

$\displaystyle = \sec \theta + 1 \text{ = RHS. } \text{ Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 48: } \frac{\cos \theta . \cot \theta}{1 + \sin \theta} = \mathrm{cosec} \theta - 1$

Answer:

$\displaystyle \text{LHS } = \frac{\cos \theta . \cot \theta}{1 + \sin \theta}$

$\displaystyle = \frac{\cos^2 \theta}{\sin \theta (1 + \sin \theta)}$

$\displaystyle = \frac{(1-\sin \theta)(1 + \sin \theta)}{\sin \theta (1 + \sin \theta)}$

$\displaystyle = \frac{1 - \sin \theta}{\sin \theta}$

$\displaystyle = \mathrm{cosec} \theta - 1 \text{ = RHS. } \text{ Hence proved. }$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

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Class 10: Trigonometry – Exercise 23(a)

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