Question

7sin²thitha+3cos²thitha=4 then tan thitha=
a)√2
b)2
c)√3
d)3​

Answer 1

EXPLANATION.

⇒ 7sin²θ + 3cos²θ = 4.

As we know that,

Divide both sides by cos²θ, we get.

⇒ 7sin²θ/(cos²θ) + 3cos²θ/(cos²θ) = 4/(cos²θ).

⇒ 7tan²θ + 3 = 4sec²θ.

As we know that,

Formula of :

⇒ 1 + tan²θ = sec²θ.

Using this formula in the equation, we get.

⇒ 7tan²θ + 3 = 4(1 + tan²θ).

⇒ 7tan²θ + 3 = 4 + 4tan²θ.

⇒ 7tan²θ - 4tan²θ = 4 - 3.

⇒ 3tan²θ = 1.

⇒ tan²θ = 1/3.

⇒ tanθ = √1/3.

⇒ tanθ = ± 1/√3.

                                                                                                                     

MORE INFORMATION.

(1) sin2θ = 2sinθcosθ = 2tanθ/1 + tan²θ.

(2) cos2θ = cos²θ - sin²θ = 2cos²θ - 1 = 1 - 2sin²θ = 1 - tan²θ/1 + tan²θ.

(3) tan2θ = 2tanθ/1 - tan²θ.

(4) sin3θ = 3sinθ - 4sin³θ.

(5) cos3θ = 4cos³θ - 3cosθ.

(6) tan3θ = 3tanθ - tan³θ/1 - 3tan²θ.

Answer 2

7sin²θ + 3cos²θ = 4
7sin²θ + 3(1-sin²θ) = 4
7sin²θ + 3 - 3sin²θ = 4
4sin²θ = 1
sin²θ = 1/4
sinθ = 1/2 ⇒ θ = 30°

.: tanθ = tan30°= 1/√3