prove:1+cosA +sinA/1+cosA-sinA = 1+sinA/cosA
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Answer:
We have,
1+csosA−sinA
1+cosA+sinA
=
1+cosA−sinA
1+cosA+sinA
×
(1+cosA)+sinA
(1+cosA)+sinA
=
(1+cosA)
2
−sin
2
A
((1+cosA)+sinA)
2
=
1+cos
2
A+2cosA−1+cos
2
A
(1+cosA)
2
+sin
2
A+2(1+cosA)sinA
=
2cos
2
A+2cosA
1+cos
2
A+sin
2
A+2cosA+2sinA+2sinAcosA
=
2cos
2
A+2cosA
1+cos
2
A+sin
2
A+2cosA+2sinA+2sinAcosA
2cosA(1+cosA)
1+1+2cosA+2sinA+2sinAcosA
2cosA(1+cosA)
2+2cosA+2sinA+2sinAcosA
cosA(1+cosA)
1+cosA+sinA+sinAcosA
=
cosA(1+cosA)
1+sinA+cosA(1+sinA)
=
cosA(1+cosA)
(1+sinA)(1+cosA)
=
cosA
1+sinA
Step-by-step explanation:
hope it's helpful to you ☺️
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