Question

prove that:
1+sec a/sec a=sin^2a/1-cos a​

Answer 1

GIVEN:

★$\dfrac{1+secA}{secA}=\dfrac{sin^{2}A}{1-cos A}$

TO PROVE:

★LHS=RHS

CONCEPT USED:

★We would first simplify RHS and LHS and try to obtain same value.

ANSWER:

LHS

=$\dfrac{1+secA}{secA}$

=$\dfrac{1}{secA}+\dfrac{\cancel{secA}}{\cancel{secA}}$

=$cosA+1$

$\large\green{\boxed{cosA=\dfrac{1}{secA}}}$

_____________________________________

RHS

=$\dfrac{sin^{2}A}{1-cos A}$

=$\dfrac{1-cos^{2}A}{1-cosA}$

$\large\purple{\boxed{(a+b)(a-b)=a^{2}-b^{2}}}$

=$\dfrac{(1+cosA)\cancel{(1-cosA)}}{\cancel{1-cosA}}$

=$1+cosA$

Hence RHS = LHS

Hence proved

Answer 2

Answer:

Step-by-step explanation:

According to this question :

Given that,

1+sec a/sec a = sin 2a/1-cos a,

Let, L.H.S,

1+sec a/sec a,

1+(1/cos a)//1 / cos a,

cos a + 1 / cos a//1/cos a,

cos a +1,

Multiply and divide with (1 - cos a),

(1+cos a)*1-cos a/1-cos a,

1-cos^2 a/1-cos a,

sin 2a/1-cos a,

R.H.S,

So,L.H.S=R.H.S,1+sec a/sec a=sin^2a/1-cos a​ is proved.

Hope it helps you.