Question

If theta≠ nπ and tan theta is the geometric mean between sin theta and cos theta , then prove that,
2-4 sin²theta + 3sin⁴ theta - sin^6 theta = 1.

Answer 1

i hope this would help

Answer 2

Answer:

$2-4sin^{2}\theta+3sin^{4}\theta-sin^{6}\theta=1$

Step-by-step explanation:

tan$\theta$ is the geometrical mean with sin$\theta$ and cos$\theta$

$tan\theta=\sqrt{sin\theta\times cos\theta}$

$tan^{2}\theta=sin\theta \times \cos\theta$

$sin\theta=cos^{3} \theta$  → (1)

Now, consider the LHS of the given expression

LHS = $2-4sin^{2}\theta+3sin^{4}\theta-sin^{6}\theta$

       = $1-sin^{2}\theta+1-sin^{6}\theta-3sin^{2}\theta+3sin^{4}\theta$

       = $1-sin^{2}\theta+(1-sin^{2}\theta )^{3}$

       = $1-sin^{2}\theta+(cos^{2}\theta )^{3}$

       = $1-sin^{2}\theta+(cos^{3}\theta )^{2}$

       = $1-sin^{2}\theta+sin^{2}\theta$   (from equation 1)

        = 1

        = RHS

Therefore,

$2-4sin^{2}\theta+3sin^{4}\theta-sin^{6}\theta=1$