Question

Prove that \[{\tan \phi \over a+\tan^2\phi} = \sin \phi \cos \phi\]

Answer 1

sorry, a = 1

Answer 2

\[1+\tan ^{2}\phi=\sec ^{2}\phi\]
so,
\[\tan \phi/(1+\tan ^{2}\phi)=\tan \phi/\sec ^{2}\phi\]
\[\tan \phi \times \cos ^{2}\phi\]
=\[(\sin \phi/\cos \phi)\times \cos ^{2}\phi\]
=\[\sin \phi \cos \phi\]

Answer 3

whys is \[\tan \phi \over \sec^2 \phi\] resulting from the other equation?

Answer 4

Because I have replaced
\[1+\tan ^{2}\phi \]
in the original question with
\[\sec ^{2}\phi\]
They are equal to each other, isnt it?

Answer 5

Ok, how does it become \[\tan \phi \times \cos^2 \phi~?\]

Answer 6

\[1/\sec ^{2}\phi=\cos ^{2}\phi\]
Again replacement of 1/sec^2phi with cos^2phi

Answer 7

(tanx)/(1+tan^(2)x) = sinxcosx

Convert the sec^(2)x to its (sin)/(cos) equivalent.
\[\huge \frac{(tanx)}{\frac{1}{\cos^2(x)}}\]

Remove the single term factors from the expression.
tanxcos^(2)(x)

Replace tanx with an equivalent expression (sinx)/(cosx) using the fundamental identities.
(sinxcos^(2)(x))/(cosx)

Reduce the expression (sinxcos(x)^(2))/(cos(x)) by removing a factor of cos(x) from the numerator and denominator
=sinxcos(x)

Answer 8

Oh, Ok.