Question

If $0^{\circ }<A<{90}^{\circ },$ then find the value of $(1+{\tan }^{2}A)(1-{\sin }^{2}A).$

• A. $0.5$
• B. $1$
• C. $-0.5$
• D. $-1$

Answer 1

The correct option is B $1$

Given: $0^{\circ }<A<{90}^{\circ }$ and we have to find the value of $(1+{\tan }^{2}A)(1-{\sin }^{2}A)$
Using the Trigonometric Identities
${\sin }^2\theta +{\cos }^2\theta =1\&1+{\tan }^2\theta ={\sec }^2\theta$
The given expression can be rewritten as:
$(1+{\tan }^{2}A)(1-{\sin }^{2}A)={\sec }^{2}A\cdot {\cos }^{2}A$

$\because \cos A=\frac{1}{\sec A}$

$\Rightarrow {\sec }^{2}A.{\cos }^{2}A=1$