43
A teacher asked his students to draw a right triangle $ABC$ with $AB = 8$ cm, $\angle B = 90^\circ$ and $BC = 15$ cm. Based on the above, answer the following :
(i) Evaluate $(\sin^2 A - \cos^2 A)$
(ii) Evaluate $(\frac{1}{\cos^2 A} - \frac{1}{\cot^2 A})$
(iii) (a) Evaluate $\frac{2 \tan A}{1 + \tan^2 A}$ and prove that it is equal to $2 \sin A \cos A$.
OR
(iii) (b) Evaluate : $\frac{\tan^2 A - \sec^2 A}{\cot^2 A - \csc^2 A}$.
(i) Evaluate $(\sin^2 A - \cos^2 A)$
(ii) Evaluate $(\frac{1}{\cos^2 A} - \frac{1}{\cot^2 A})$
(iii) (a) Evaluate $\frac{2 \tan A}{1 + \tan^2 A}$ and prove that it is equal to $2 \sin A \cos A$.
OR
(iii) (b) Evaluate : $\frac{\tan^2 A - \sec^2 A}{\cot^2 A - \csc^2 A}$.
Show SolutionHide Solution↓
(i) Hypotenuse $AC = 17$ cm $\implies \sin^2 A - \cos^2 A = (\frac{15}{17})^2 - (\frac{8}{17})^2 = \frac{161}{289}$
(ii) $\frac{1}{\cos^2 A} - \frac{1}{\cot^2 A} = \frac{17^2}{8^2} - \frac{15^2}{8^2} = 1$
(iii) (a) $\frac{2 \tan A}{1 + \tan^2 A} = \frac{2 \times \frac{15}{8}}{1 + \frac{15^2}{8^2}} = \frac{240}{289}$
$2 \sin A \cos A = 2 \times \frac{15}{17} \times \frac{8}{17} = \frac{240}{289}$ Hence they are equal
Note : Marks should be awarded to the alternate solution as well : $\frac{2 \tan A}{\sec^2 A} = \frac{2 \times \frac{\sin A}{\cos A}}{\frac{1}{\cos^2 A}} = 2 \sin A \cos A$
OR
(iii) (b) $\frac{\tan^2 A - \sec^2 A}{\cot^2 A - \csc^2 A} = \frac{(\frac{15}{8})^2 - (\frac{17}{8})^2}{(\frac{8}{15})^2 - (\frac{17}{15})^2} = 1$
(ii) $\frac{1}{\cos^2 A} - \frac{1}{\cot^2 A} = \frac{17^2}{8^2} - \frac{15^2}{8^2} = 1$
(iii) (a) $\frac{2 \tan A}{1 + \tan^2 A} = \frac{2 \times \frac{15}{8}}{1 + \frac{15^2}{8^2}} = \frac{240}{289}$
$2 \sin A \cos A = 2 \times \frac{15}{17} \times \frac{8}{17} = \frac{240}{289}$ Hence they are equal
Note : Marks should be awarded to the alternate solution as well : $\frac{2 \tan A}{\sec^2 A} = \frac{2 \times \frac{\sin A}{\cos A}}{\frac{1}{\cos^2 A}} = 2 \sin A \cos A$
OR
(iii) (b) $\frac{\tan^2 A - \sec^2 A}{\cot^2 A - \csc^2 A} = \frac{(\frac{15}{8})^2 - (\frac{17}{8})^2}{(\frac{8}{15})^2 - (\frac{17}{15})^2} = 1$