Math, asked by anantrajusharma, 10 months ago

please solve it fast ...


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Answers

Answered by Nikkisely
4

Answer:

tan(A+B)/cot(A-B)=tan^2A-tan^2B/tan^Atan^B

LHS:

=tan(A+B)×tan(A-B)

=tanA+TanB/1-tanAtanB × tanA-tanB/1-tanAtanB

=tan^A-tan^B/1-tan^Atan^B

=RHS

Hope it helps u

.........

and muje aapse koi dushmani Nahi hai,,

Good day:-)

Answered by Anonymous
20

AnswEr :

Given that,

 \sf \:  \dfrac{ tan(a   +  b) }{ cot(a - b) }  =  \dfrac{ tan {}^{2} (a) -  { tan }^{2}(b)  }{1 - {tan}^{2}a. {tan}^{2} b  }

LHS :

 \sf \:  \dfrac{ tan(a   +  b) }{ cot(a - b) }

We Know That,

tan∅ = 1/cot∅

Thus,

 \longrightarrow \:  \sf \:  tan( a + b) .tan(a - b)

Now,

 \sf \:  tan(a + b)  =  \dfrac{ tan(a)  +  tan(b) }{1 - tan(a)tan(b)}

Also,

 \sf \: tan(a - b) =  \dfrac{tan(a) - tan(b)}{1 + tan(a)tan(b)}

Thus,

 \longrightarrow \:  \sf \:  \bigg(  \dfrac{tan(a) + tan(b)}{1 - tan(a)tan(b)} \bigg) \bigg(  \dfrac{tan(a) - tan(b)}{1 + tan(a)tan(b)} \bigg)

We know that,

(a + b)(a - b) = a² - b²

 \longrightarrow \:  \sf \:  \dfrac{ {tan}^{2} (a) -  {tan}^{2}(b) }{1 -  {tan}^{2} (a)tan {}^{2}(b) }

Hence, ProveD

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