Math, asked by rumajana9898, 4 months ago

OR
Evaluate :
(sec? 27° - cot? 63°) + (sin52° + sin’38°
(cosec 34º – tan56°) + tan 10°.tan 20°.tan 30°. tan 70°. tan 80°​

Answers

Answered by neetuagarwal4191
0

Answer:

We have, csc257o−tan233osec254o−cot236o+2sin238o.sec252o−sin245o.

=csc2(90o−33o)−tan233osec2(90o−36o)−cot236o+2sin238o.sec2(90o−38o)−sin245o

=sec233o−tan233ocsc236o−cot236o+2sin238o.csc2

Step-by-step explanation:

We have, csc257o−tan233osec254o−cot236o+2sin238o.sec252o−sin245o.

=csc2(90o−33o)−tan233osec2(90o−36o)−cot236o+2sin238o.sec2(90o−38o)−sin245o

=sec233o−tan233ocsc236o−cot236o+2sin238o.csc2

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