Question

prove that: ( see the pic)
step by step ​

Answer 1

Step-by-step explanation:

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plz refer to pic

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Answer 2

HEY MATE YOUR ANSWER IS HERE...

BY TAKING LHS

$\frac{sin \: a - sin \: b \: }{cos \: a + cos \: b} + \frac{cos \: a - cos \: b}{sin \: a +sin \: b }$

$= \frac{(sin \: a - sin \: b)(sin \: a + sin \: b) + (cos \: a + cos \: b)(cos \: a - cos \: b)}{(sin \: a + sin \: b)(cos \: a + cos \: b)}$

NOW ,

BY IDENTITY

a²-b² = ( a + b )( a - b )

$= \frac{ {sin}^{2} a - {sin}^{2} b + {cos}^{2}a - {cos}^{2} b }{(sin \: a + sin \: b)(cos \: a + cos \: b)}$

NOW BY REARRANGING TERMS

$= \frac{ ({sin}^{2} a + {cos}^{2}a) - ({cos}^{2} b + {sin}^{2} b ) }{(sin \: a + sin \: b)(cos \: a + cos \: b)}$

NOW BY TRIGNOMATRIC IDENTITY

SIN²∅ + COS²∅ = 1

HENCE ,

$= \frac{ 1 - 1 }{(sin \: a + sin \: b)(cos \: a + cos \: b)}$

$= \frac{ 0 }{(sin \: a + sin \: b)(cos \: a + cos \: b)}$

HENCE = 0

HENCE PROVED

THANKS FOR UR QUESTION HOPE IT HELPS

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