Question

PROVE THAT :-
CosA/1-TanA + SinA/1-CotA = SinA+CosA​

Please Explain it Step by Step, I feel Trigonometry as a hard chapter for me.

Answer 1

EXPLANATION.

$\sf \implies \dfrac{cos(A)}{1 - tan(A)} \ + \ \dfrac{sin(A)}{1 - cot(A)} = sin(A) + cos(A).$

As we know that,

Formula of :

⇒ tanθ = sinθ/cosθ.

⇒ cotθ = cosθ/sinθ.

Using this formula in equation, we get.

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

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

$\sf \implies \dfrac{cos^{2} (A)}{cos(A) - sin(A)} \ + \ \dfrac{sin^{2} (A)}{sin(A) - cos(A)}$

$\sf \implies \dfrac{cos^{2} (A)}{cos(A) - sin(A)} \ - \ \dfrac{sin^{2} (A)}{cos(A) - sin(A)}$

$\sf \implies \dfrac{cos^{2}(A) - sin^{2} (A) }{cos(A) - sin(A)}$

$\sf \implies \dfrac{[cos(A) - sin(A)][cos(A) + sin(A)]}{cos(A) - sin(A)}$

$\sf \implies cos(A) + sin(A).$

                                                                                                                       

MORE INFORMATION.

Fundamental trigonometric identities.

(1) = sin²θ + cos²θ = 1.

(2) = 1 + tan²θ = sec²θ.

(3) = 1 + cot²θ = cosec²θ.

Answer 2

Given -

PROVE THAT :-

CosA/1-TanA + SinA/1-CotA = SinA+CosA

Required Answer prove

$\frac{ \cos(a) }{1 - \tan(a) } + \frac{ \sin(a) }{1 - \cot(a) } \\ \\ \frac{ \cos(a) } { \frac { 1 - \sin(a) }{ \cos(a) } } + \frac{ \sin(a) }{ \frac{ \cos(a) } { 1 - \sin(a) } } \\ \\ \\ \frac{ { \cos(a) }^{2}}{ \cos(a) - \sin(a) } - \frac{ \sin(a) }{ \tan(a) }$