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प्रश्न
For triangle ABC, show that : `tan (B + C)/2 = cot A/2`
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उत्तर
We know that for a triangle ΔABC
∠A + ∠B + ∠C = 180°
∠B + ∠C = 180° – ∠A
`=> (angle B + angle C)/2 = 90^circ - (angle A)/2`
`=> tan ((B + C)/2) = tan (90^circ - A/2)`
= `cot (A/2)`
संबंधित प्रश्न
If the angle θ= –60º, find the value of cosθ.
Write all the other trigonometric ratios of ∠A in terms of sec A.
Prove the following trigonometric identities.
(secθ + cosθ) (secθ − cosθ) = tan2θ + sin2θ
Evaluate:
`sin80^circ/(cos10^circ) + sin59^circ sec31^circ`
If tanθ = 2, find the values of other trigonometric ratios.
If A + B = 90°, then \[\frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A}\]
In the following Figure. AD = 4 cm, BD = 3 cm and CB = 12 cm, find the cot θ.
Prove that:
\[\frac{sin\theta \cos(90° - \theta)cos\theta}{\sin(90° - \theta)} + \frac{cos\theta \sin(90° - \theta)sin\theta}{\cos(90° - \theta)}\]
Evaluate: `3(sin72°)/(cos18°) - (sec32°)/("cosec"58°)`.
Prove the following:
tan θ + tan (90° – θ) = sec θ sec (90° – θ)
