Math, asked by Anonymous, 9 months ago

2tan^2 60°+4 sin^2 60-2tan ^2 30°× sec^2 30°÷2 sin^2 45° - 5 sin 90°+3 cos ^2 90°-6cos 60°

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Answered by chnageswarr
1

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Answered by anubaskota2
0

Answer:

2tan^2 60°+4 sin^2 60-2tan ^2 30°× sec^2 30°÷2 sin^2 45° - 5 sin 90°+3 cos ^2 90°-6cos 60°

​or,2( √ 3)^2+2( √ 3/2)^2-2(1/ √ 3)^2*( √ 3)^2/2(1/ √ 2)^2-5*1+3*0-6*1/2

6+3-2/(1/2)-5+0-6

-6

Step-by-step explanation:

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