Question

$\sqrt{1+ta{n}^2\theta }\sqrt{1+co{t}^2\theta }\sqrt{1-co{s}^2\theta }\sqrt{1-si{n}^2\theta }=$

• A. 1
• B. $sin\theta$
• C. $cos\theta$
• D. $sec\theta$

Answer 1

The correct option is A 1

W know that,
$si{n}^2\theta +co{s}^2\theta =1,1+ta{n}^2\theta =se{c}^2\theta ,1+co{t}^2\theta =cose{c}^2\theta$

$\sqrt{1+ta{n}^2\theta }\times \sqrt{1+co{t}^2\theta }$
$\times \sqrt{1-co{s}^2\theta }\times \sqrt{1-si{n}^2\theta }$
$=\sqrt{se{c}^2\theta }\times \sqrt{cose{c}^2\theta }\times$
$\sqrt{si{n}^2\theta }\times \sqrt{co{s}^2\theta }$
$=sec\theta \times cosec\theta \times sin\theta \times cos\theta$
$=(sec\theta \times cos\theta )\times (cosec\theta \times sin\theta )=1\times 1=1$