Question

1) iv. If 7Sin^2theta + 3 Cos^2theta = 4 then prove that tan theta = 1 / √3

Class 10

Content Quality Solution Required

❎ Don't Spamming ❎

Answer 1

7.sin²θ + 3.cos²θ = 4

7.sin²θ + ౩( 1 - sin²θ ) = 4

4.sin²θ + 3 = 4 ; 4.sin²θ = 1

sin²θ = 1/4

sin θ = + 1/2 ; θ = π/6 ; θ = π/3

Therefore , tan θ = 1/√3

Answer 2

Method - 1:

Given : 7sin^2A + 3cos^2A = 4.

= > 7 sin^2A + 3cos^2A = 4 * 1

= > 7sin^2A + 3cos^2A = 4(sin^2A + cos^2A)

= > 7sin^2A + 3cos^2A = 4sin^2A + 4cos^2A

= > 7sin^2A - 4sin^2A = 4cos^2A - 3cos^2A

= > 3sin^2A = cos^2A

= > 3 sin^2A = cos^2A

= > cos^2A/sin^2A = 3.

= > cot^2A = 3

$= > cotA = \sqrt{3}$

We know that tanA = (1/cotA)

$= > tanA = \frac{1}{\sqrt{3}}$

----------------------------------------------------------------------------------------------------------------

Method 2:

$Given : tanA = \frac{1}{\sqrt{3}}$

$= > tanA = tan(30)$

= > A = 30.

Now,

$= > 7sin^2A + 3cos^2A = 4$

$= > 7(\frac{1}{4}) + 3(\frac{3}{4} ) = 4$

$= > \frac{7}{4} + \frac{9}{4} = 4$

$= > \frac{16}{4} = 4$

$= > 4 = 4.$

---------------------------------------------------------------------------------------------------------------

Hope this helps!