Simplify:- tan 2 60 0 + 4 cos 2 45 0 + 3( sec 2 30 0 + cos 2 90 0 )
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Answered by
8
tan² 60° + 4cos²45° +3[sec²30° +cos²90°]
√3 + 4(1/√2) +3[(2/√3)+0]
we get the following eqn
(√3)² + 4(1/√2)² +3[(2/√3)²+(0)²]
3+4/2+6/3
we get
3+2+2
⇒ 7
hope it's hep u
√3 + 4(1/√2) +3[(2/√3)+0]
we get the following eqn
(√3)² + 4(1/√2)² +3[(2/√3)²+(0)²]
3+4/2+6/3
we get
3+2+2
⇒ 7
hope it's hep u
Answered by
6
=√3 + 4(1/√2) + 3{(2√3) + 0}
Now,
=(√3)^2 + 4(1/√2) ^2+3{(2√3)^2+0
=3 + 4(1/2) + 3(4/3) + 0
=3+2+4
=9
Step-by-step explanation:
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