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. $\sqrt{1+ta{n}^2\theta }$
• B. $\sqrt{1+se{c}^2\theta }$
• C. $cos\theta$
• D. $\sqrt{1+si{n}^2\theta }$

Answer 1

The correct option is A

$\sqrt{1+ta{n}^2\theta }$

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

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

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

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

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

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

= $sec\theta$

= $\sqrt{1+ta{n}^2\theta }$