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प्रश्न
Without using a trigonometric table, prove that
`(cos 70°)/(sin 20°) + (cos 59°)/(sin 31°) - 8sin^2 30° = 0`.
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उत्तर
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.
संबंधित प्रश्न
Prove the following trigonometric identities. `(1 - cos A)/(1 + cos A) = (cot A - cosec A)^2`
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
\[\frac{\sin \theta}{1 + \cos \theta}\]is equal to
Prove that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.
Prove that: `1/(sec θ - tan θ) = sec θ + tan θ`.
Prove the following identities.
cot θ + tan θ = sec θ cosec θ
`(1 - tan^2 45^circ)/(1 + tan^2 45^circ)` = ?
Prove that cot2θ × sec2θ = cot2θ + 1
If 2 cos θ + sin θ = `1(θ ≠ π/2)`, then 7 cos θ + 6 sin θ is equal to ______.
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`
