Question

Which of the following has the same value as the given expression?

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

• A. √(1+tan²θ)
• B. √(1+sec²θ)
• C. cosθ
• D. √(1+sin²θ)

Answer 1

The correct option is A

√(1+tan²θ)

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

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

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

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

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

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

= $sec\theta$

= $\sqrt{1+ta{n}^2\theta }$ = R.H.S