Prove the following

We know that,

$\tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}$

$\Rightarrow \tan \theta=\frac{\cos \alpha\left(\frac{\sin \alpha}{\cos \alpha}-1\right)}{\cos \alpha\left(\frac{\sin \alpha}{\cos \alpha}+1\right)}$

Since, $\tan \mathrm{A}=(\sin \mathrm{A}) /(\cos \mathrm{A})$

$\Rightarrow \tan \theta=(\tan \alpha-1) /(\tan \alpha+1)$

Since, $\tan \pi / 4=1$

$\Rightarrow \tan \theta=\frac{\left(\tan \alpha-\tan \frac{\pi}{4}\right)}{\left(1+\tan \frac{\pi}{4} \cdot \tan \alpha\right)}$

We know that,

tan(x-y) = (tan x – tan y) / (1 + tan x . tan y)

Therefore, tan θ = tan ( α – π/4)

⇒ θ = α – π/4

⇒ α = θ + π/4 …(i)

To prove,

sinα + cosα = √2 cos θ

∵ LHS = sinα + cosα

From equation (i)

⇒ LHS = sin(θ + π/4) + cos(θ + π/4)

∵ sin(x + y) = sin x cos y + cos x sin y

And, cos(x + y) = cos x cos y – sin x sin y

Therefore, LHS = sin θ cos(π/4) + sin(π/4)cos θ + cos θ cos(π/4) – sin(π/4)sin θ

∵ sin(π/4)=cos(π/4) = 1/√2

⇒ LHS = sin θ (1/√2) + (1/√2) cos θ + cos θ (1/√2) – sin θ (1/√2)

⇒ LHS = 2 cos θ (1/√2)

⇒ LHS = √2 cos θ = RHS

Therefore, sinα + cosα = √2 cos θ