Question

It tan θ = 1/√3, then 7 sin²θ + 3 cos²θ =

A) 16/4

B) 7/4

C) 9/4

D) 1

Answer 1

Correct option is: A) \(\frac {16}4\)

We have tan \(\theta\) = \(\frac 1{\sqrt3} = tan \, 30^\circ\)

 = \(\theta\) = \(30^\circ\)

Now \(7\, sin^2\theta + 3\, cos^2\theta = 7\, sin^2\,30^\circ + 3\, cos^2\, 30^\circ\).

= 7 \(\times (\frac 12)^2 + 3 (\frac {\sqrt3}{2})^2 \) (\(\because\) sin \(30^\circ\) = \(\frac 12\) & cos \(30^\circ\) = \(\frac {\sqrt3}2\).

= \(\frac 74 + 3 \times \frac 34 = \frac 74 + \frac 94 = \frac {16}4 = 4\)

Alternative method :

We have tan \(\theta\) = \(\frac 1{\sqrt3} \)

\(\therefore\) \(Sec^2\theta = 1 + tan^2\theta = 1 + (\frac 1{\sqrt3})^2 = 1+ \frac 13 = \frac 43\)

Now, \(7\, sin^2\theta + 3\, cos^2\theta = \frac {7\, sin^2\theta + 3\, cos^2\theta}{cos^2\theta}.cos^2\theta\).

= \(\frac {(7\, tan^2\theta + 3)}{sec^2\theta}\) (\(\because\) \(\frac 1{cos\theta} = sec \theta\))

= \(\frac {7\times \frac 13 + 3}{\frac 43}\)

= \(\frac {\frac {7+9}{3}}{\frac 43} = \frac {16}4 = 4\).

Answer 2

Correct option is: A) \(\frac{16}{4}\)