(1 + x²) cosĂ = 4x then Prove it that
Co secA
+
Cot A =1+2x
\1-2x
Answers
Step-by-step explanation:
Given that
(1+4x^{2})cosA=4x(1+4x
2
)cosA=4x
We can rearrange the above equation as
cosA=\frac{4x}{(1+4x^{2})}cosA=
(1+4x
2
)
4x
... (1)
As we know that
cosA=\frac{Base}{Hyp}cosA=
Hyp
Base
... (2)
Comparing (1) and (2)
Base=4xBase=4x
Hyp=(1+4x^{2})Hyp=(1+4x
2
)
Now by pythagorean theorem we can find that
Perp=1-4x^{2}Perp=1−4x
2
As we know
cosecA=\frac{hyp}{perp}cosecA=
perp
hyp
cosecA=\frac{1+4x^{2}}{1-4x^{2}}cosecA=
1−4x
2
1+4x
2
cotA=\frac{base}{perp}cotA=
perp
base
cosecA=\frac{4x}{1-4x^{2}}cosecA=
1−4x
2
4x
Now
cosecA + cotA = \frac{1+4x^{2}}{1-4x^{2}}+\frac{4x}{1-4x^{2}}
1−4x
2
1+4x
2
+
1−4x
2
4x
cosecA + cotA = \frac{1+4x^{2} + 4x}{1-4x^{2}}
1−4x
2
1+4x
2
+4x
cosecA + cotA = \frac{(1 + 2x)^{2}}{(1-2x)(1+2x)}
(1−2x)(1+2x)
(1+2x)
2
cancelling the same terms in numerator and denominator
cosecA + cotA = \frac{(1 + 2x)}{(1-2x)}
(1−2x)
(1+2x)
Hence proved