Question

If 7sin square theta + 3cos square theta=4 show that tan theta = 1/ root3

Answer 1

Solution is very easy
$let \: \alpha = theta \\ 7 \sin {}^{2} ( \alpha ) + 3 \cos { }^{2} ( \alpha ) = 4 \\ 7(1 - \cos {}^{2} ( \alpha ) ) + 3 \cos {}^{2} ( \alpha ) = 4 \\ 7 - 4 \cos {}^{2} ( \alpha ) = 4 \\ \cos {}^{2} ( \alpha ) = \frac{3}{4} \\ \cos( \alpha ) = \frac{ \sqrt{3} }{2 } \\ \alpha = {30}^{0} \\ \tan( \alpha ) = \tan( {30}^{0 } ) \\ \tan( {30}^{0} ) = \frac{1}{ \sqrt{3 } } \\ hence \: proved$
Hence Proved

Answer 2

Step-by-step explanation:

Answer :-

→ tan30° = 1/√3

Step-by-step explanation :-

We have,

→ 7 sin² ∅ + 3 cos² ∅ = 4 .

⇒ 4 sin²∅ + 3 sin²∅ + 3 cos²∅ = 4 .

⇒ 4 sin²∅ + 3( sin²∅ + cos²∅ ) = 4 .

⇒ 4 sin²∅ + 3( 1 ) = 4 . [ ∵ sin²∅ + cos²∅ = 1 ] .

⇒ 4 sin²∅ + 3 = 4 .

⇒ 4 sin²∅ = 4 - 3 .

⇒ 4 sin²∅ = 1 .

⇒ sin²∅ = 1/4 .

⇒ sin ∅ = √(1/4) .

∴ sin ∅ = 1/2 .

But, sin 30° = 1/2 .

Then, sin ∅ = sin 30° .

$\huge \pink{ \boxed{ \it \therefore \theta = 30 \degree.}}$

Then, tan 30° = 1/√3 . .....

Hence, it is proved .