Question

Class 10
Topic - Trigonometry

Prove that -

$\rm \dfrac{cos \theta}{(1-tan \theta)}+ \dfrac{sin \theta}{(1- cot \theta)} = (cos \theta + sin \theta)$

Answer 1

Answer:

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Answer 2

Answer:

Given that -

$\rm \dfrac{cos \theta}{(1-tan \theta)}+ \dfrac{sin \theta}{(1- cot \theta)} = (cos \theta + sin \theta)$

Consider LHS :

⇒ $\rm \dfrac{cos \theta}{(1-tan \theta)}+ \dfrac{sin \theta}{(1- cot \theta)}$

We know that -

• tanθ = sinθ/cosθ
• cotθ = cosθ/sinθ

⇒ $\rm \dfrac{cos \theta}{(1- \dfrac{\sin \theta}{\cos \theta})}+ \dfrac{sin \theta}{(1- \dfrac{\cos \theta}{\sin \theta})}$

⇒ $\rm \dfrac{cos \theta}{(\dfrac{\cos \theta - \sin \theta}{\cos \theta})}+ \dfrac{sin \theta}{( \dfrac{\sin \theta - \cos \theta}{\sin \theta})}$

⇒ $\rm \dfrac{\cos^2 \theta}{(\cos \theta - \sin \theta)} + \dfrac{\sin^2 \theta}{(\sin \theta - \cos \theta)}$

⇒ $\rm \dfrac{\cos^2 \theta}{(\cos \theta - \sin \theta)} - \dfrac{\sin^2 \theta}{(\cos \theta - \sin \theta)}$

⇒ $\rm \dfrac{\cos^2 \theta - \sin^2 \theta }{(\cos \theta - \sin \theta)}$

⇒ $\rm \dfrac{(\cos \theta - \sin \theta)(\cos \theta + \sin \theta)}{(\cos \theta - \sin \theta)}$

⇒ $\cos \theta + \sin \theta = \text{RHS}$

∴ LHS = RHS

Hence, it is proved.