Question

PLS HELP!!! URGENT. I NEED THE ANSWER AND AM NOT GETTING IT.​

Answer 1

Answer:

here theta is x, i didnt got theta symbol

tan x=√3 . sin x

1/cos x=√3

cos x = 1/√3

cos²x=1/3

cos²x=1-sin²x

so,

1/3=1-sin²x

sin²x=2/3

sin²x-cos²x=(2/3) -(1/3) =1/3

sorry for incovinience

hope u understood

mark me as brainliest

Answer 2

Given :

(√3) tan θ = 3 sin θ

To find :

Value of

sin²θ - cos²θ

Identity used :

• tan x = sin x ÷ cos x
• sin²x + cos²x = 1

Solution :

(√3) tan θ = 3 sin θ

$\sf \implies \: \sqrt{3} \: \times \dfrac{ \sin( \theta) }{ \cos( \theta) } \: = \: 3 \times \sin( \theta) \\ \\ \sf \implies \: \sqrt{3} \: \times \dfrac{ \sin( \theta) }{ 3 \times \sin( \theta) \: } \: = \: \cos( \theta) \\ \\ \sf \implies \: \cos( \theta) \: = \: \dfrac{ \sqrt{3} }{3}$

Squaring both side

$\sf \implies \: \cos^2( \theta) \: = \:\bigg( \dfrac{ \sqrt{3} }{3} \bigg)^2 \\ \\ \sf \implies \: \cos^2( \theta) \: = \:\dfrac{ {( \sqrt{3}) }^{2} }{3^2} \\ \\ \sf \implies \: \cos^2( \theta) \: = \:\dfrac{ 3 }{9}$

$\sf \implies \: \cos^2( \theta) \: = \:\dfrac{ 1}{3} \: \: \: \: \: \: equation1$

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Now , we know that ,

sin²θ + cos²θ = 1

sin²θ = 1 - cos²θ

On putting value of cos²θ from equation 1 into above equation, we get;

$\sf \implies \: \sin^2( \theta) \: = \: 1 - \dfrac{1}{3} \\ \\ \sf \: take \: lcm \\ \sf \implies \: \sin^2( \theta) \: = \: \dfrac{(1 \times 3) - (1 \times 1)}{3} \\ \\ \sf \implies \: \sin^2( \theta) \: = \: \dfrac{3 - 1}{3} \\ \\ \sf \implies \: \sin^2( \theta) \: = \: \dfrac{2}{3}$

$\sf \implies \: \sin^2( \theta) \: = \: \dfrac{2}{3} \: \: \: \: equation \: 2$

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Now put value of sin²θ & cos²θ in sin²θ - cos²θ .

$\sf \implies \: \sin^2( \theta) \: - \: \cos^2( \theta) \: = \: \: \dfrac{2}{3} \: - \: \dfrac{ 1}{3} \\ \\ \sf \implies \: \sin^2( \theta) \: - \: \cos^2( \theta) \: = \: \: \dfrac{2 - 1}{3} \: \\ \\ \sf \implies \: \sin^2( \theta) \: - \: \cos^2( \theta) \: = \: \: \bold{\dfrac{1}{3}} \:$

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ANSWER : (1/3)