Advertisements
Advertisements
प्रश्न
If tanA = `(1 - cos "B")/sin"B"`, then tan2A = ______.
Advertisements
उत्तर
Given that: tanA = `(1 - cos "B")/sin"B"`
tan2A = `(2tan"A")/(1 - tan^2"A")`
= `(2((1 - cos "B")/sin"B"))/(1 - ((1 - cos "B")/sin"B")^2`
= `(2((2sin^2 "B"/2)/(2sin "B"/2 cos "B"/2)))/(1 - ((2sin^2 "B"/2)/(2sin "B"/2 cos "B"/2))^2` ........`[(because 1 - cos "B" = 2sin^2 "B"/2),(sin"B" = 2sin"B"/2 cos"B"/2)]`
= `(2((sin "B"/2)/(cos "B"/2)))/(1 - ((sin "B"/2)/(cos "B"/2))^2`
= `(2tan "B"/2)/(1 - tan^2 "B"/2)`
= tanB
APPEARS IN
संबंधित प्रश्न
Prove that: \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} = \tan x\]
Prove that: \[\sqrt{2 + \sqrt{2 + 2 \cos 4x}} = 2 \text{ cos } x\]
Prove that: \[\frac{\cos 2 x}{1 + \sin 2 x} = \tan \left( \frac{\pi}{4} - x \right)\]
Show that: \[2 \left( \sin^6 x + \cos^6 x \right) - 3 \left( \sin^4 x + \cos^4 x \right) + 1 = 0\]
Prove that: \[\cos^6 A - \sin^6 A = \cos 2A\left( 1 - \frac{1}{4} \sin^2 2A \right)\]
Prove that: \[\cos^6 A - \sin^6 A = \cos 2A\left( 1 - \frac{1}{4} \sin^2 2A \right)\]
Prove that: \[\cot \frac{\pi}{8} = \sqrt{2} + 1\]
If \[\cos x = - \frac{3}{5}\] and x lies in the IIIrd quadrant, find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2}, \sin 2x\] .
If \[\text{ tan } x = \frac{b}{a}\] , then find the value of \[\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}\] .
If \[\tan A = \frac{1}{7}\] and \[\tan B = \frac{1}{3}\] , show that cos 2A = sin 4B
If \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha + \cos \beta = b\] , prove that
(ii) \[\cos \left( \alpha - \beta \right) = \frac{a^2 + b^2 - 2}{2}\]
If \[\cos x = \frac{\cos \alpha + \cos \beta}{1 + \cos \alpha \cos \beta}\] , prove that \[\tan\frac{x}{2} = \pm \tan\frac{\alpha}{2}\tan\frac{\beta}{2}\]
Prove that: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]
Prove that: \[\cos 36° \cos 42° \cos 60° \cos 78° = \frac{1}{16}\]
Prove that: \[\cos\frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{5\pi}{15} \cos\frac{6\pi}{15} \cos \frac{7\pi}{15} = \frac{1}{128}\]
If \[\pi < x < \frac{3\pi}{2}\], then write the value of \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}\] .
If \[\text{ tan } A = \frac{1 - \text{ cos } B}{\text{ sin } B}\]
, then find the value of tan2A.
If \[\text{ sin } x + \text{ cos } x = a\], then find the value of
If \[\text{ sin } x + \text{ cos } x = a\], find the value of \[\left|\text { sin } x - \text{ cos } x \right|\] .
If \[\cos 2x + 2 \cos x = 1\] then, \[\left( 2 - \cos^2 x \right) \sin^2 x\] is equal to
If in a \[∆ ABC, \tan A + \tan B + \tan C = 0\], then
If \[2 \tan \alpha = 3 \tan \beta, \text{ then } \tan \left( \alpha - \beta \right) =\]
The value of \[\tan x \sin \left( \frac{\pi}{2} + x \right) \cos \left( \frac{\pi}{2} - x \right)\]
If \[A = 2 \sin^2 x - \cos 2x\] , then A lies in the interval
The value of \[\cos^2 \left( \frac{\pi}{6} + x \right) - \sin^2 \left( \frac{\pi}{6} - x \right)\] is
\[2 \left( 1 - 2 \sin^2 7x \right) \sin 3x\] is equal to
If \[\tan \frac{x}{2} = \frac{\sqrt{1 - e}}{1 + e} \tan \frac{\alpha}{2}\] , then \[\cos \alpha =\]
The value of \[\cos^4 x + \sin^4 x - 6 \cos^2 x \sin^2 x\] is
The value of \[\tan x \tan \left( \frac{\pi}{3} - x \right) \tan \left( \frac{\pi}{3} + x \right)\] is
If \[n = 1, 2, 3, . . . , \text{ then } \cos \alpha \cos 2 \alpha \cos 4 \alpha . . . \cos 2^{n - 1} \alpha\] is equal to
If A = cos2θ + sin4θ for all values of θ, then prove that `3/4` ≤ A ≤ 1.
Prove that sin 4A = 4sinA cos3A – 4 cosA sin3A
The value of `sin pi/18 + sin pi/9 + sin (2pi)/9 + sin (5pi)/18` is given by ______.
