Advertisements
Advertisements
प्रश्न
Prove that:
tan 20° tan 40° tan 80° = `sqrt3`.
उत्तर
tan 20° tan 40° tan 80°
`= (sin 20^circ)/(cos 20^circ) xx (sin 40^circ)/(cos 40^circ) xx (sin 80^circ)/(cos 80^circ)`
`= (sin 20^circ xx sin 40^circ xx sin 80^circ)/(cos 20^circ cos 40^circ cos 80^circ)`
Consider sin 20° × sin 40° sin 80°
= sin 20° sin (60° – 20°) sin (60° + 20°)
= sin 20° [sin2 60° – sin2 20°]
`= sin 20^circ [3/4 - sin^2 20^circ]`
`= sin 20^circ [(3 - 4 sin^2 20^circ)/4]`
`= (3 sin 20^circ - 4 sin^3 20^circ)/4`
`= (sin 60^circ)/4`
`= (sqrt3/2)/4 = sqrt3/8`
cos 20° × cos 40° cos 80° = `1/8` ....[∵ from (i)] .... (2)
divide (1) by (2) we get, tan 20° tan 40° tan 80° = `(sqrt3/8)/(1/8) = sqrt3`
APPEARS IN
संबंधित प्रश्न
Show that:
sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0
Prove that:
sin 50° + sin 10° = cos 20°
Prove that:
Prove that:
If cos (α + β) sin (γ + δ) = cos (α − β) sin (γ − δ), prove that cot α cot β cot γ = cot δ
If A + B = \[\frac{\pi}{3}\] and cos A + cos B = 1, then find the value of cos \[\frac{A - B}{2}\].
Write the value of \[\sin\frac{\pi}{15}\sin\frac{4\pi}{15}\sin\frac{3\pi}{10}\]
If cos (A + B) sin (C − D) = cos (A − B) sin (C + D), then write the value of tan A tan B tan C.
If sin (B + C − A), sin (C + A − B), sin (A + B − C) are in A.P., then cot A, cot B and cot Care in
If \[\tan\alpha = \frac{x}{x + 1}\] and