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brilliants [131]
3 years ago
12

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

Mathematics
1 answer:
Virty [35]3 years ago
7 0

Answer:

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

Step-by-step 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\times \tan \theta \times \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 \times \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.

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