Question

Which of the following options is equal to the given expresssion?

$\frac{cot({90}^{\circ }-\theta )}{cose{c}^2\theta }\times \frac{sec\theta .co{t}^3\theta }{si{n}^2({90}^{\circ }-\theta )}$

• A. $tan\theta$
• B. $sec\theta$
• C. $cos\theta$
• D. $\sqrt{1+se{c}^2\theta }$

Answer 1

The correct option is B $sec\theta$

Given: $\frac{cot({90}^{\circ }-\theta )}{cose{c}^2\theta }\times \frac{sec\theta .co{t}^3\theta }{si{n}^2({90}^{\circ }-\theta )}$

$cot({90}^{\circ }-\theta )=tan\theta$

$sin\left(\right({90}^{\circ }-\theta )=cos\theta$

$\Rightarrow \frac{tan\theta }{cose{c}^2\theta }$ $\times$ $\frac{sec\theta co{t}^3\theta }{co{s}^2\theta }$

= $tan\theta$$\times$ $\frac{si{n}^2\theta }{co{s}^2\theta }$ $\times$ $sec\theta$ $\times$$co{t}^3\theta$

= $tan\theta$ $\times$ $ta{n}^2\theta$ $\times$ $sec\theta$ $\times$ $co{t}^3\theta$

= $ta{n}^3\theta$ $\times$ $co{t}^3\theta$ $\times$ $sec\theta$

= $sec\theta$