Question

6
If sino + coso = V2coso, (0 # 90°) then the value of tand is
a) V2 - 1
b) V2 + 1
c) √2
d) -V2​

Answer 1

The required "option a) $\sqrt{2}-1$" is correct.

Step-by-step explanation:

We have,

$\sin \theta+\cos \theta=\sqrt{2}\cos \theta$

where, $\theta\neq 0$

To find, the value of $\tan \theta = ?$

∴ $\sin \theta+\cos \theta=\sqrt{2}\cos \theta$

Dividing both sides by $\cos \theta,$ we get

$\dfrac{\sin \theta+\cos \theta}{\cos \theta} =\dfrac{\sqrt{2}\cos \theta}{\cos \theta}$

⇒  $\dfrac{\sin \theta}{\cos \theta}+ \dfrac{\cos \theta}{\cos \theta}=\sqrt{2}$

⇒ $\tan \theta+1=\sqrt{2}$

⇒ $\tan \theta=\sqrt{2}-1$

∴ The value of $\tan \theta=\sqrt{2}-1$

Thus, the required "option a) $\sqrt{2}-1$" is correct.