Question

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PROVE THAT :
(secA - cosA )( cotA + tanA ) = tanA×secA



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Answer 1

Gd Eve :)

LHS, =(SecA-CosA)(CotA+TanA) =(1/CosA-CosA)(CosA/SinA+SinA/CosA) =((1-Cos^2)/CosA)((Cos^2A+Sin^2A)/SinACosA) =(Sin^2A/CosA)(1/SinACosA) =(SinASinA/SinACosACosA) =(SinA/CosACosA) =(SinA/CosA)(1/CosA) =TanASecA=RHS }Thus proved

Answer 2

HELLO!

_____________________________

$\mathbf{(secA-cosA)(cotA+tanA)=tanA xsecA}$

$\mathbf{<u>L</u><u>H</u><u>S</u><u>:</u>}$

= (secA-cosA)(cotA+tanA)

= (secA-$\frac{1}{secA}$)($\frac{1}{tanA}$+tanA)

= (sec²A-1/secA)(1+tan²A/tanA)

= (tan²A/secA)(sec²A/tanA)

$\mathbf{= tanA x secA}$

$\mathbf{<u>R</u><u>H</u><u>S</u>}$

= tanA x secA

Here,

$\boxed{LHS = RHS}$

$\mathbf{Hence Proved!}$