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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
Evaluate sin25° cos65° + cos25° sin65°
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x = 3 cosec θ + 4 cot θ
y = 4 cosec θ – 3 cot θ
If sec2 θ (1 + sin θ) (1 − sin θ) = k, then find the value of k.
The value of (1 + cot θ − cosec θ) (1 + tan θ + sec θ) is
If x = a cos θ and y = b sin θ, then b2x2 + a2y2 =
If x = h + a cos θ, y = k + b sin θ.
Prove that `((x - h)/a)^2 + ((y - k)/b)^2 = 1`.
`5/(sin^2θ) - 5cot^2θ`, complete the activity given below.
Activity:
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= `square (1/(sin^2θ) - cot^2θ)`
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= 5(1)
= `square`
