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Question
If A + B = 90°, show that `(sin B + cos A)/sin A = 2tan B + tan A.`
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Solution
LHS = `(sin B + sec A)/sin A`
= `(sin (90 - A) + sec A)/sin A`
= `(cos A + sec A)/sin A`
= `(cos^2 A + 1)/(sin A. cos A)`
= `(2cos^2 A + sin^2 A)/(sin A. cos A)`
= 2cot A + tan A
= 2 tan B + tan A = RHS
hence proved.
RELATED QUESTIONS
Prove the following trigonometric identities.
if `T_n = sin^n theta + cos^n theta`, prove that `(T_3 - T_5)/T_1 = (T_5 - T_7)/T_3`
If `cos theta = 2/3 , "write the value of" ((sec theta -1))/((sec theta +1))`
If sin θ − cos θ = 0 then the value of sin4θ + cos4θ
Prove the following identity :
`2(sin^6θ + cos^6θ) - 3(sin^4θ + cos^4θ) + 1 = 0`
Prove the following identity :
`(sec^2θ - sin^2θ)/tan^2θ = cosec^2θ - cos^2θ`
Prove that `sqrt((1 - sin θ)/(1 + sin θ)) = sec θ - tan θ`.
Prove that `( tan A + sec A - 1)/(tan A - sec A + 1) = (1 + sin A)/cos A`.
If x = h + a cos θ, y = k + b sin θ.
Prove that `((x - h)/a)^2 + ((y - k)/b)^2 = 1`.
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
Prove the following identity:
(sin2θ – 1)(tan2θ + 1) + 1 = 0
