Math, asked by Arshad2004, 11 months ago

Prove that :-
(cot³A.sin³A)/(cosA+sinA)² + (tan³A.cos³A)/(cosA+sinA)² = (secA.cosecA-1)/(cosecA+secA)

need the answer ASAP

Answers

Answered by priyaverma2558
2

Step-by-step explanation:

LHS

 \frac{ { \cos( \alpha ) }^{3} }{( { \cos( \alpha )  +  \sin( \alpha )) }^{2} }  +  \frac{ { \sin( \alpha ) }^{3} }{( { \sin( \alpha )  +  \cos( \alpha ) )}^{2} }

We have formula ,

a³ +b³ = a+b(a² -ab+b²)

 \frac{ \cos( \alpha )  +  \sin( \alpha )( \sin {}^{2} ( \alpha ) -  \sin( \alpha )   \cos( \alpha )  +  cos {}^{2} ( \alpha )   }{( \sin( \alpha ) +  \cos( \alpha )  ){}^{2}  }

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

Important steps -

(a+b)² = (a+b)(a+b) (step 3)

.

sin ²A +cos ²A = 1 (step 3)

tan A = sin A/ cos A ( used in step 1)

cot A = cos A / sin A

RHS - Steps

Cosec A = 1/sin A

Sec A = 1/ cos A

 \frac{1 -  \sin( \alpha )  \cos( \alpha ) }{ \sin( \alpha ) \cos( \alpha )  }  \times  \frac{ \sin( \alpha )  \cos( \alpha ) }{ \cos( \alpha )  +  \sin( \alpha ) }

 \frac{1 -  \sin( \alpha ) \cos( \alpha )  }{ \sin( \alpha )  +  \cos( \alpha ) }

Hence proved

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