Math, asked by apinderathi, 1 year ago

Simplify:- tan 2 60 0 + 4 cos 2 45 0 + 3( sec 2 30 0 + cos 2 90 0 )

Answers

Answered by Anonymous
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

Answered by mukherjeesania19
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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