Question

If sinθ + cosθ = √2cosθ, (θ ≠ 90°) then the value of tanθ is a) √2 − 1 b) √2 + 1 c) √2 d) −√2

Answer 1

Answer:

Hence option a) √2-1 is correct.

tanθ= √2-1

Step-by-step explanation:

To find the value of tanθ,

If sinθ + cosθ = √2cosθ, (θ ≠ 90°)

We know that

$tan \theta = \frac{sin \theta}{cos \theta}$

To do the same ,divide the LHS and RHS of equation by cos θ

$\frac{sin \theta}{cos \theta} + \frac{cos\theta}{cos \theta} = \sqrt{2} \frac{cos \theta}{cos \theta} \\ \\ tan \theta + 1 = \sqrt{2} \\ \\ tan \theta = \sqrt{2} - 1$

Hence option a) √2-1 is correct.

Hope it helps you.

Answer 2

Hence option a) √2-1 is correct.

tanθ= √2-1

Step-by-step explanation:

To find the value of tanθ,

If sinθ + cosθ = √2cosθ, (θ ≠ 90°)

We know that

$tan \theta = \frac{sin \theta}{cos \theta}$

To do the same ,divide the LHS and RHS of equation by cos θ

$\frac{sin \theta}{cos \theta} + \frac{cos\theta}{cos \theta} = \sqrt{2} \frac{cos \theta}{cos \theta} \\ \\ tan \theta + 1 = \sqrt{2} \\ \\ tan \theta = \sqrt{2} - 1$

Hence option a) √2-1 is correct.

Hope it helps you.