Question

If 7 sin^2 θ + 3 cos^2 θ = 4, find the value of √3tan θ.

Answer 1

7sin²θ + 3cos²θ = 4

7sin²θ + ౩(1-sin²θ) = 4

$7sin²θ+3-3sin²θ = 4$

$4sin²θ+3 = 4$

$4sin²θ=1$

$sin²θ = 1/4$

$sinθ = +1/2$

$sinθ = +  π/6$

$Therefore \: \: \: \: tan θ = + π/6$

then we get tan θ = + 1/√3

 

As the required answer is  tan θ = + 1/√3 (here we considered   θ =+ π/6)

Answer 2

Answer:

7sin²θ+3cos²θ=4

7sin²θ + ౩(1-sin²θ) = 47sin²θ+౩(1−sin²θ)=4

7sin²θ+3-3sin²θ = 47sin²θ+3−3sin²θ=4

4sin²θ+3 = 44sin²θ+3=4

4sin²θ=14sin²θ=1

sin²θ = 1/4sin²θ=1/4

sinθ = +1/2sinθ=+1/2

sinθ = + π/6sinθ= + π/6

Therefore \: \: \: \: tan θ = + π/6Thereforetan θ=+ π/6

then \: \: \: we \: \: \: get \: \: \: tan θ = + 1/√3thenwegettan θ= +1/√3

As \: the \: required \: answer \: \: is \: tan θ = + 1/√3 (here \: \: we \: \: considered \: \: θ =+ π/6)Astherequiredansweris tan θ=+1/√3(hereweconsidered θ=+ π/6)

Step-by-step explanation:

thx please