Question

Solve:
$4[1–si{n}^2\theta ][1+ta{n}^2\theta ][1+cose{c}^2\theta ]$

• A. $4\times co{t}^2\theta$
• B. $4\times ta{n}^2\theta$
• C. $3\times si{n}^2\theta$
• D. $3\times co{t}^2\theta$

Answer 1

The correct option is A $4\times co{t}^2\theta$

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

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

$4[1-si{n}^2\theta ][1+ta{n}^2\theta ][1+cose{c}^2\theta ]=4\times co{s}^2\theta \times se{c}^2\theta \times co{t}^2\theta =4\times co{t}^2\theta$