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Question
Without using a trigonometric table, prove that
`(cos 70°)/(sin 20°) + (cos 59°)/(sin 31°) - 8sin^2 30° = 0`.
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Solution
We have,
LHS = `(cos 70°)/(sin 20°) + (cos 59°)/(sin 31°) - 8sin^2 30° = 0`.
= `cos(90° - 20°)/(sin 20°) + cos(90° - 31°)/(sin 31°) - 8 xx (1/2)^2`
= `(sin 20°)/(sin 20°) + (sin 31°) /(sin 31°) - 8 xx 1/4`
= 1 + 1 - 2
= 2 -2
= 0
= RHS
Hence proved.
RELATED QUESTIONS
Prove the following trigonometric identities.
`cot theta - tan theta = (2 cos^2 theta - 1)/(sin theta cos theta)`
Prove the following identities:
sec2 A . cosec2 A = tan2 A + cot2 A + 2
Prove the following identities:
`cosA/(1 - sinA) = sec A + tan A`
Prove that:
`1/(cosA + sinA - 1) + 1/(cosA + sinA + 1) = cosecA + secA`
Prove the following identities:
sec4 A (1 – sin4 A) – 2 tan2 A = 1
If `( cos theta + sin theta) = sqrt(2) sin theta , " prove that " ( sin theta - cos theta ) = sqrt(2) cos theta`
Define an identity.
If x = r sin θ cos ϕ, y = r sin θ sin ϕ and z = r cos θ, then
Prove that cot2θ – tan2θ = cosec2θ – sec2θ.
Find the value of sin2θ + cos2θ
Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
