Math, asked by sanamqwer123, 8 hours ago

( tanβ+ cotβ ) ( tan²β - tanβ . cotβ + cot²β)

Answers

Answered by Kaushalsingh74883508

4

Step-by-step explanation:

We need to prove cotβ−2tan(α−β)

Thus cotβ−2[

1+tanα.tanβ

tanα−tanβ

]=0

Taking LHS:

1−tanα.tanβ

1

[cotβ+tanα−2tanα+2tanβ]

1−tanα.tanβ

1

[cotβ−tanα+2tanβ]

=0

=RHS

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