Question

sinA/1-cosA+tanA/1+cosA=secA.cosecA+cotA​

Answer 1

$\huge{\mathfrak{Solution}}$

Given:-

$\bold{\frac{sinA}{1-cosA}+\frac{tanA}{1+cosA}=secA.\;cosecA+cotA}$

We take L.H.S.

$\bold{\implies \frac{sinA}{1-cosA}+\frac{tanA}{1+cosA}}$

$\bold{\implies \frac{sinA(1+cosA)+tanA(1-cosA)}{1-cos^{2}A}}$

$\bold{\implies \frac{sinA+sinA\;cosA+tanA-tanA\;cosA}{sin^{2}A}}$

$\bold{\implies \frac{1}{sinA}+\frac{cosA}{sinA}+\frac{1}{cosA\;sinA}-\frac{1}{sinA}}$

$\bold{\implies \frac{cosA}{sinA}+\frac{1}{cosA\;sinA}}$

$\bold{\implies cotA+secA\;cosecA = RHS}$

$\bold{Hence\;Proved}$

Some Trigonometric Identities:-

$\bold{1).\; sin(90^{\circ}-A) = cosA}$

$\bold{2).\;tan(90^{\circ}-A)=cotA}$

$\bold{3).\;sec(90^{\circ}-A) = cosecA}$

$\bold{4).\;cos^{2}A+sin^{2}A=1}$

$\bold{5).\;1+tan^{2}A=sec^{2}A}$

$\bold{6).\;cot^{2}A+1=cosec^{2}A}$

Answer 2

L.H.S

=Sin A(1+ cos A)+tan A(1- cos A)/1- cos^2 A

=Sin A + Sin A. cos A +tan A -tan A.cos A/sin^2 A

=Sin A +Sin A. Cos A+tan A - sinA /Cos A + cos A/sin^2 A

=Sin A +Sin A.cos A + tan A-Sin A/Sin^2 A

=Sin A. cos A +sin A/Cos A/sin^2 A

=Sin A(cos^2 A +1/cos A)/sin^2 A

=Cos^2 A+1/cos A. sinA

=Cos^2 A/cos A. SinA +1/cos A. sinA

=cos A/Sin A +Sec A.Cosec A= Cot A + Sec A .Cosec A