Question

(tan theta+ cot theta)^2= sec^2 theta+cosec^2 theta​

Answer 1

Proved that
`(\tan \theta + \cot \theta)^(2) = \sec^(2) \theta + \csc^(2) \theta`.

Explanation:

We have to prove that

`(\tan \theta + \cot \theta)^(2) = \sec^(2) \theta + \csc^(2) \theta`

Now, Left hand side

=
`(\tan \theta + \cot \theta)^(2)`

=
`\tan^(2) \theta + 2* \tan \theta * \cot \theta + \cot^(2) \theta` {Since we know the formula (a + b)² = a² + 2ab + b²}

=
`\tan^(2) \theta + 2 + \cot^(2) \theta`

{Since
`\tan \theta * \cot \theta = 1`}

=
`(\tan^(2) \theta + 1) + (\cot^(2) \theta + 1)`

=
`\sec^(2) \theta + \csc^(2) \theta`

{Since we know the identities:

`\sec^(2) x = \tan^(2) x + 1` and
`\csc^(2) x = \cot^(2) x + 1`}

= Right hand side

Hence, proved.