1/sinA+cosA - 1/sinA-cosA
Answers
Answer:
Proof :
Taking LHS,
LHS=\frac{1+\sin A-\cos A}{1+\sin A+\cos A}LHS=
1+sinA+cosA
1+sinA−cosA
Using the trigonometry identities we can write,
1) \sin A=2\sin (\frac{A}{2})\cos (\frac{A}{2})sinA=2sin(
2
A
)cos(
2
A
)
2) \cos A=2\cos^2(\frac{A}{2})-1cosA=2cos
2
(
2
A
)−1
3) \cos A=1-2\sin^2(\frac{A}{2})cosA=1−2sin
2
(
2
A
)
Applying these identities,
LHS=\frac{1+2\sin (\frac{A}{2})\cos (\frac{A}{2})-1+2\sin^2(\frac{A}{2})}{1+2\sin (\frac{A}{2})\cos (\frac{A}{2})+2\cos^2(\frac{A}{2})-1}LHS=
1+2sin(
2
A
)cos(
2
A
)+2cos
2
(
2
A
)−1
1+2sin(
2
A
)cos(
2
A
)−1+2sin
2
(
2
A
)
LHS=\frac{2\sin (\frac{A}{2})\cos (\frac{A}{2})+2\sin^2(\frac{A}{2})}{2\sin (\frac{A}{2})\cos (\frac{A}{2})+2\cos^2(\frac{A}{2})}LHS=
2sin(
2
A
)cos(
2
A
)+2cos
2
(
2
A
)
2sin(
2
A
)cos(
2
A
)+2sin
2
(
2
A
)
LHS=\frac{2\sin (\frac{A}{2})(\cos (\frac{A}{2})+\sin(\frac{A}{2}))}{2\cos (\frac{A}{2})(\sin (\frac{A}{2})+\cos(\frac{A}{2}))}LHS=
2cos(
2
A
)(sin(
2
A
)+cos(
2
A
))
2sin(
2
A
)(cos(
2
A
)+sin(
2
A
))
Cancel the like term,
LHS=\frac{\sin (\frac{A}{2})}{\cos (\frac{A}{2})}LHS=
cos(
2
A
)
sin(
2
A
)
LHS=\tan \frac{A}{2}LHS=tan
2
A
LHS=RHSLHS=RHS
Hence proved.
Step-by-step explanation:
1+sinA+cosA
1+sinA−cosA
=tan(
2
A
)
Proof :
Taking LHS,
LHS=\frac{1+\sin A-\cos A}{1+\sin A+\cos A}LHS=
1+sinA+cosA
1+sinA−cosA
Using the trigonometry identities we can write,
1) \sin A=2\sin (\frac{A}{2})\cos (\frac{A}{2})sinA=2sin(
2
A
)cos(
2
A
)
2) \cos A=2\cos^2(\frac{A}{2})-1cosA=2cos
2
(
2
A
)−1
3) \cos A=1-2\sin^2(\frac{A}{2})cosA=1−2sin
2
(
2
A
)
Applying these identities,
LHS=\frac{1+2\sin (\frac{A}{2})\cos (\frac{A}{2})-1+2\sin^2(\frac{A}{2})}{1+2\sin (\frac{A}{2})\cos (\frac{A}{2})+2\cos^2(\frac{A}{2})-1}LHS=
1+2sin(
2
A
)cos(
2
A
)+2cos
2
(
2
A
)−1
1+2sin(
2
A
)cos(
2
A
)−1+2sin
2
(
2
A
)
LHS=\frac{2\sin (\frac{A}{2})\cos (\frac{A}{2})+2\sin^2(\frac{A}{2})}{2\sin (\frac{A}{2})\cos (\frac{A}{2})+2\cos^2(\frac{A}{2})}LHS=
2sin(
2
A
)cos(
2
A
)+2cos
2
(
2
A
)
2sin(
2
A
)cos(
2
A
)+2sin
2
(
2
A
)
LHS=\frac{2\sin (\frac{A}{2})(\cos (\frac{A}{2})+\sin(\frac{A}{2}))}{2\cos (\frac{A}{2})(\sin (\frac{A}{2})+\cos(\frac{A}{2}))}LHS=
2cos(
2
A
)(sin(
2
A
)+cos(
2
A
))
2sin(
2
A
)(cos(
2
A
)+sin(
2
A
))
Cancel the like term,
LHS=\frac{\sin (\frac{A}{2})}{\cos (\frac{A}{2})}LHS=
cos(
2
A
)
sin(
2
A
)
LHS=\tan \frac{A}{2}LHS=tan
2
A
LHS=RHSLHS=RHS