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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
What is the value of (cos2 67° – sin2 23°)?
Prove that:
`(cos(90^circ - theta)costheta)/cottheta = 1 - cos^2theta`
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Prove that:
sin (28° + A) = cos (62° – A)
If 3 cot θ = 4, find the value of \[\frac{4 \cos \theta - \sin \theta}{2 \cos \theta + \sin \theta}\]
Prove that:
cos15° cos35° cosec55° cos60° cosec75° = \[\frac{1}{2}\]
Find the value of the following:
tan 15° tan 30° tan 45° tan 60° tan 75°
The value of the expression (cos2 23° – sin2 67°) is positive.
