Math, asked by tusharpanjwani1506, 2 months ago

[tan³Φ/1+tan²Φ]+
[cot³Φ/1+cot²Φ]=
secΦ×cosecΦ-2sinΦcosΦ

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Answers

Answered by ushadubey555

1

Answer:

tan³Φ/1+tan²Φ]+

[cot³Φ/1+cot²Φ]=

secΦ×cosecΦ-tan³Φ/1+tan²Φ]+

[cot³Φ/1+cot²Φ]=

secΦ×cosecΦ-2sinΦcosΦ

Step-by-step explanation:

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