Math, asked by shashankumar1410, 3 months ago

3 sin 28° sec 62° + cos2 20 ° + cos2 70 ° − sec2 45° + sin 30°​

Answers

Answered by parulkalita2
1

Answer:

3sin28° sec62°+cos^20°+cos^70°-sec^45°+sin30°

=3sin(90°-62°) sec62° +cos^(90°-70°) +cos^70°-sec^45°+1/2

=3cos62°sec62°+sin^70°+cos^70°-sec^45°+1/2

=3+1-sec^45°+1/2

=4+1/2-sec^45°

=2-sec^45°

=2-(√2) ^

=2-2

=0

Answered by harshkatara2005
0

Answer:

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