Question

The trigonometric expression ${\cot }^{2}0\left[\frac{\sec \theta -1}{1+\sin \theta }\right]+{\sec }^{2}0\left[\frac{\sin \theta -1}{1+\sec \theta }\right]$ has the value

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is B $0$

${\cot }^{2}0\left[\frac{\sec \theta -1}{1+\sin \theta }\right]+{\sec }^{2}0\left[\frac{\sin \theta -1}{1+\sec \theta }\right]$

$=\frac{co{t}^2\theta (sec\theta -1)(1-sec\theta )+se{c}^2\theta (sin\theta -1)(1+sin\theta )}{(1+sin\theta )(1+sec\theta )}$

$=\frac{co{t}^2\theta (se{c}^2\theta -1)+se{c}^2\theta (si{n}^2\theta -1)}{(1+sin\theta )(1+sec\theta )}$

$=\frac{co{t}^2\theta ta{n}^2\theta +se{c}^2\theta (-co{s}^2\theta )}{(1-sin\theta )(1+sec\theta )}$

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

$=0$

identity used $:-se{c}^2\theta -1=ta{n}^2\theta$

$1-si{n}^2\theta =co{s}^2\theta$