Prove the following: sinx- sinycosx+cosy=tan x-y2 - Mathematics

Question

Prove the following:

$\frac{\sin x - \sin y}{\cos x + \cos y} = \tan \frac{x - y}{2}$

Sum

Solution

We have, बायाँ पक्ष = $\frac{\sin x - \sin y}{\cos x + \cos y}$

= $\frac{2 \sin (\frac{x - y}{2}) \cos (\frac{x + y}{2})}{2 \cos (\frac{x - y}{2}) \cos (\frac{x + y}{2})}$

= $\frac{\frac{\sin (x - y)}{2}}{\frac{\cos (x - y)}{2}}$

= tan $\frac{x - y}{2}$ = दायाँ पक्ष।

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Trigonometric Functions of Sum and Difference of Two Angles

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Q 18 Q 17 Q 19

Chapter 3: Trigonometric Functions - Exercise 3.3 [Page 73]

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NCERT Mathematics [English] Class 11

Chapter 3 Trigonometric Functions
Exercise 3.3 | Q 18 | Page 73

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Prove the following: sinx- sinycosx+cosy=tan x-y2 - Mathematics

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