Math, asked by somyatridarsinidash, 8 months ago

√2 tan ∆ =√b find the value of sin ∆+√3cos∆​

Answers

Answered by raghavpaltanwale
1

Step-by-step explanation:

for now let us replace ∆ with A

Given: √2 tanA= √b

To find: sinA+ √3cosA

√2 tanA= √b (given)

tanA=√b/√2

tanA=Opposite/Adjacent

let, opposite=√bx and adjacent=√2x

hence,

hypotenuse²=(√bx)²+(√2x)²

hypotenuse²=(b+2)x²

hypotenuse=√(b+2)x

hence,

SinA=opposite/hypotenuse

sinA=√bx/√(b+2)x

sinA=√b/√(b+2)

cosA=Adjacent/Hypotenuse

cosA=√2x/√(b+2)x

cosA=√2/√(b+2)

√3cosA=√3×√2/√(b+2)

√3cosA=√6/√(b+2)

hence,

sinA+√3cosA=√b/√(b+2)+√3/√(b+2)

sinA+√3cosA=(√b+√3)/√(b+2)

hence, replacing A with the required symbol, we have,

sin∆+√3cos∆=(√b+√3)/√(b+2)

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