(sinA/1-cosA -1-cosA/sinA)(cosA/1-sinA- 1-sinA/cosA)
Answers
Answer:
Correct Question :-
Prove that : sinA/(1 - cosA) + tanA/(1 + cosA) = (cos²A + 1)/sinA * cos A
Solution :-
\sf \dfrac{sinA}{1 - cosA} + \dfrac{tanA}{1 + cosA} = \dfrac{cos^{2} A + 1}{sinA \times cosA}
1−cosA
sinA
+
1+cosA
tanA
=
sinA×cosA
cos
2
A+1
Consider RHS
\sf = \dfrac{sinA}{1 - cosA} + \dfrac{tanA}{1 + cosA}=
1−cosA
sinA
+
1+cosA
tanA
Taking LCM
\sf = \dfrac{sinA(1 + cosA) + tanA(1 - cosA}{(1 - cosA)(1 + cosA)}=
(1−cosA)(1+cosA)
sinA(1+cosA)+tanA(1−cosA
\sf = \dfrac{sinA+ sinA.cosA + tanA - tan A.cosA}{ {1}^{2} - cos ^{2} A}=
1
2
−cos
2
A
sinA+sinA.cosA+tanA−tanA.cosA
[ Because (x - y)(x + y) = x² - y² ]
\sf = \dfrac{sinA+ sinA.cosA + tanA - \dfrac{sinA}{cosA} .cosA}{1 - cos ^{2} A}=
1−cos
2
A
sinA+sinA.cosA+tanA−
cosA
sinA
.cosA
[ Because tanA = sinA/cosA ]
\sf = \dfrac{sinA+ sinA.cosA + tanA - sinA}{sin^{2} A}=
sin
2
A
sinA+sinA.cosA+tanA−sinA
\sf = \dfrac{sinA.cosA + tanA}{sin^{2} A}=
sin
2
A
sinA.cosA+tanA
\sf = \dfrac{sinA.cosA + \dfrac{sinA}{cosA} }{sin^{2} A}=
sin
2
A
sinA.cosA+
cosA
sinA
Taking sinA common in numerator
\sf = \dfrac{sinA \bigg(cosA + \dfrac{1}{cosA} \bigg)}{sin^{2} A}=
sin
2
A
sinA(cosA+
cosA
1
)
\sf = \dfrac{ cosA + \dfrac{1}{cosA}}{sin A}=
sinA
cosA+
cosA
1
Taking LCM in numerator
\sf = \dfrac{ \dfrac{cos^{2} A +1 }{cosA}}{sin A}=
sinA
cosA
cos
2
A+1
\sf = \dfrac{cos^{2} A +1 }{sinA.cosA}=
sinA.cosA
cos
2
A+1
Hence proved
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