Question

1+sinA/ 1-sinA = secA + tanA​

Answer 1

Answer:

this is your answer buddy

Answer 2

Answer:

L.H.S= 1+SINA/1-SINA

R.H.S=(SECA+TANA)^2

lets solve right hand side::

We know secA=1/cosA and tanA =sinA/cosA::

Putiing these values in the R.H.S we get::

(1/cosA + sinA/cosA)^2

We can easily add since the denominator(cosA) is same for both the terms and get R.H.S =

(1 + sinA/cosA)^2

Now if we apply the value of square, the eq. will not get any messier as it would hv earlier:: so after squaring we get R.H.S =

(1+sinA)^2 ÷ cos^2A.(eq.1)

We know that (a+b)^2 can be written as (a+b)×(a+b)::

Hence (1+csinA)^2= (1+sinA)×(1+sinA)(eq.2)

And also cos^A= 1-sin^2A= (1^2 - sin^2A)

Which is similar to the equation (a+b)×(a-b)=a^2 - b^2.

So the term becomes (1+sinA)×(1-sinA) (eq.3)

Now put values of eq.2 and eq.3 in eq.1::

R.H.S=(1+sinA)×(1+sinA)÷(1-sinA)(1+sinA)::

Now cancel out the similar terms and we get the answer:::

hope this helps you.

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