Question

please answer question 7 about class 11 trigonometry

Answer 1

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

Applying Componendo and dividendo rule

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

$tan(\frac{\pi }{4}+\theta) = tan \alpha$

$\frac{\pi }{4} + \theta = \alpha$

$RHS = \sqrt{2}cos \theta\\\\= \sqrt{2}cos(\alpha - \frac{\pi }{4})\\ \\= \sqrt{2}(cos\alpha \times cos\frac{\pi }{4} + sin\alpha \times sin\frac{\pi }{4})\\\\=\sqrt{2}(cos\alpha \times \frac{1 }{\sqrt{2} } + sin\alpha \times\frac{1}{\sqrt{2} })\\\\= \sqrt{2} \times \frac{1}{\sqrt{2} } (cos\alpha + sin\alpha )\\\\=cos\alpha + sin\alpha = LHS$

HENCE PROVED