Question

If cosθ = 2√2/3, verify that tan²θ – sin²θ = tan²θ·sinθ.

Answer 1

Given, $cos\theta=\frac{2\sqrt{2}}{3}$
we know, cosine = base/hypotenuse
so, cosθ = 2√2/3 = base/hypotenuse
therefore, base = 2√2
and hypotenuse = 3
from Pythagoras theorem,
perpendicular = √(hypotenuse ² - base ²}
= √{3² - (2√2)²} = 1

now, tanθ = perpendicular/base = 1/2√2
sinθ = perpendicular/hypotenuse = 1/3

LHS = tan²θ - sin²θ
= {1/2√2}² - {1/3}²
= 1/8 - 1/9
= 1/72

RHS = tan²θ.sin²θ
= {1/2√2}² × {1/3}²
= 1/8 × 1/9
= 1/72

LHS = RHS

Answer 2

Here is correction in the right hand side as shown in the attachment.