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