prove that (1+sinA)/(1-sinA) = (secA+tenA)^2
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Answer:
How do we prove 1+sinA/1-sinA= (secA+tanA) ^2?
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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:::
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Answer:
this is not mathematically possible
check ur question again