Math, asked by nanduadi, 1 year ago

please solve this trignometic ques​

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Answers

Answered by nids007
2

=sec 29/sec(90-61)+2cot8.cot17.cot45.cot73.cot82-3sin^2 38+sin^52)

=1+2cot8.cot17.1.tan(90-73).tan(90-82)-3(sin^2 38+cos ^2(90-52)

= 1+2cot8.cot17.tan17.tan8-3(sin^2 38+cos^38)

=1+2-3=0

Hope it helps you.

Answered by ayushmaancristiano
2

Answer:

Step-by-step explanation:

sec 29/sec(90-61)+2cot8.cot17.cot45.cot73.cot82-3sin^2 38+sin^52)

=1+2cot8.cot17.1.tan(90-73).tan(90-82)-3(sin^2 38+cos ^2(90-52)

= 1+2cot8.cot17.tan17.tan8-3(sin^2 38+cos^38)

=1+2-3=0

Hope it helps you.

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