Question

If 7sin²∅ + 3cos²∅ = 4, show that tan∅ = 1/√3​

Answer 1

$\sf\orange{Given:}$

$\sf{7sin^{2}\theta+3cos^{2}\theta=4}$

$\sf\red{To \ prove:}$

$\sf{tan\theta=\frac{1}{\sqrt3}}$

$\sf\green{\underline{\underline{Proof:}}}$

$\sf{7sin^{2}\theta+3cos^{2}\theta=4}$

$\sf\blue{cos^{2}\theta=1-sin^{2}\theta}$

$\sf\blue{... Trigonometric \ identity}$

$\sf{\therefore{7sin^{2}\theta+3(1-sin^{2}\theta)=4}}$

$\sf{\therefore{7sin^{2}\theta+3-3sin^{2}\theta=4}}$

$\sf{\therefore{4sin^{2}\theta=4-3}}$

$\sf{\therefore{4sin^{2}\theta=1}}$

$\sf{\therefore{sin^{2}\theta=\frac{1}{4}}}$

$\sf{On \ taking \ square \ root \ of \ both \ sides}$

$\sf{sin\theta=\frac{1}{2}}$

$\sf{But, \ sin30°=\frac{1}{2}}$

$\sf{\therefore{\theta=30°}}$

$\sf{we \ know, \ tan30°=\frac{1}{\sqrt3}}$

$\sf{\therefore{tan\theta=\frac{1}{\sqrt3}}}$

Answer 2

$\huge\red{Answer}$

Solution :-

7sin²∅ + 3cos²∅ = 4

= = > 4sin²∅ + 3sin²∅ + 3cos²∅ = 4

= = > 4sin²∅ + 3( sin²∅ + cos²∅) = 4

= = > 4sin²∅ + 3 × 1 = 4 [sin²∅ + cos²∅ = 1]

= = > 4 sin²∅ = 1

= = > sin²∅ = 1/4

Therefore cos²∅ = (1-sin²∅) = (1-1/4) = 3/4

tan²∅ = sin²∅/cos²∅ = (1/4 × 4/3) = 1/3

Hence, tan ∅ = 1/3