Question

(sinA+cosA)(secA+cosecA)=2+secA.cosecA

prove that ..
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Answer 1

Solution :

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Given :

To prove :

(Sin A + Cos A)(Sec A + Cosec A)= 2 + Sec A Cosec A

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We know that,

⇒ $Cosec A = \frac{1}{Sin A}$

&

⇒ $Sec A = \frac{1}{Cos A}$

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Proof :

LHS = (Sin A + Cos A)(Sec A + Cosec A)= 2 + Sec A Cosec A

⇒ $(Sin A + Cos A)( \frac{1}{Cos A} + \frac{1}{Sin A} )$

⇒ $(Sin A + Cos A)( \frac{Sin A + Cos A}{Sin A Cos A} )$

⇒ $\frac{(Sin A + Cos A)(Sin A + Cos A)}{Sin A Cos A}$

⇒ $\frac{Sin^2A + 2Sin A Cos A + Cos^2 A}{Sin A Cos A}$

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We also know that,

⇒ Sin² A + Cos² A = 1

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Hence,

 ⇒ $\frac{2Sin A Cos A + 1}{Sin A Cos A}$

⇒ $\frac{2Sin A Cos A}{Sin A Cos A} + \frac{1}{Sin A Cos A}$

⇒ $2 + \frac{1}{Sin A Cos A}$

⇒ $2 + (\frac{1}{Sin A})( \frac{1}{Cos A} )$

⇒ $2 + (\frac{1}{Cos A})( \frac{1}{Sin A} )$

⇒ 2 + Sec A Cosec A

⇒ RHS

                                             ∴ LHS = RHS

                                          ∴ Hence, Proved,.

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                                         Hope it Helps !!

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Answer 2

Step-by-step explanation:

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