Math, asked by Friedman, 22 hours ago

1+cotA/cosa +1+tanA/sinA = 2(secA+cosecA)​

Answers

Answered by Aeryxz
11

★ To prove</p><p>• [tex] \sf \dfrac{1 +  \cot(A) }{ \cos(A) }  +  \dfrac{1 +  \tan(A) }{ \sin(A) }  = 2( \sec A +  \csc \: A)

★ Solution :-

 \sf \: LHS =  \dfrac{1 +  \cot(A) }{ \cos(A) }  +  \dfrac{1 +  \tan(A) }{ \sin(A) }

 \sf \implies   \dfrac{ \sin(A) (1 +  \dfrac{ \cos(A) }{ \sin(A) } ) + \cos(A) (1 +  \dfrac{ \sin(A) }{ \cos(A) } )}{ \sin(A) \cos(A)  }

 \implies \sf \:  \dfrac{ \sin(A)  +  \cos(A)  +  \cos(A) +  \sin(A)  }{\sin(A) \cos(A)}

 \implies \sf \:  \dfrac{2 \sin(A) }{\sin(A) \cos(A)}  +  \dfrac{2 \cos(A) }{\sin(A) \cos(A)}

 \implies \sf \: 2( \dfrac{1}{ \cos(A) } +  \dfrac{1}{ \sin(A) }  )

 \implies \sf \: 2( \sec(A)  +  \csc(A) ) = RHS

Hence, Proved !!

Answered by najahaven
1

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I hope you understand the answer!!

Given in two methods...

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