Question

If 3 cos theta = 1 , find the value of 6 sin square theta + 10 square theta upon 4 cos theta​

Answer 1

Answer:

3 cosθ = 1

or, cosθ = 1/3

or, b/h = 1/3

b=1, h=3

According to Pythogorean theorem,

h² = p² + b²

or, 3² = p² + 1²

p = √8 = 2√2

sinθ = p/h = $\frac{2\sqrt{2}}{3}$

tanθ = p/b = 2√2

$\frac{6sin^2\theta + tan^2\theta}{cos\theta}$ = $\frac{6*(\frac{2\sqrt{2}}{3} )^2 + (2\sqrt{2})^2}{\frac{1}{3} }$

= $\frac{\frac{16+24}{3} }{\frac{1}{3} }$

= 40