Question

$If \;sin\theta+cos\theta=\sqrt{2}, b = \dfrac{\sqrt{2}(sin\theta+cos\theta) }{sin\theta cos\theta } , find\; b?$

Answer 1

Answer:

4

Step-by-step explanation:

Given :

$\mathrm{sin\theta + cos\theta = \sqrt{2} \ ; \ b = \dfrac{\sqrt{2}(sin\theta + cos\theta)}{sin\theta cos\theta}}$

To find :

The value of b

Solution :

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

Squaring on both sides,

$\mathrm{(sin\theta + cos\theta)^2 = (\sqrt{2})^2 }$

$\implies \mathrm{sin^2\theta + 2sin\theta cos\theta + cos^2\theta = 2 }$

We know that,

$\boxed{\mathrm{sin^2x + cos^2x = 1}}$

$\implies \mathrm{1 + 2sin\theta cos\theta = 2}$

$\implies \mathrm{2sin\theta cos\theta = 2 - 1}$

$\implies \mathrm{ 2sin\theta cos\theta = 1}$

$\implies \mathrm{sin\theta cos\theta = \dfrac{1}{2}}$

So, substituting this is b, we get,

$\longmapsto b = \dfrac{\sqrt{2}(sin\theta + cos\theta)}{sin\theta cos\theta}$

$\implies \mathrm{b = \dfrac{\sqrt{2}\times\sqrt{2}}{\bigg(\dfrac{1}{2}\bigg)}}$

$\implies \mathrm{b = 2 \times 2}$

$\therefore \underline{\boxed {\mathrm{b =4}}}$

Hope it helps!!