Advertisements
Advertisements
प्रश्न
Prove that: \[\frac{\sin 2x}{1 + \cos 2x} = \tan x\]
Advertisements
उत्तर
\[LHS = \frac{\sin2x}{1 + \cos2x}\]
\[ = \frac{2\sin x \times \cos x}{1 + 2 \cos^2 x - 1} \] `[∵ sin 2 x = 2 sin x xx cos x and cos 2x = 2 cos ^2 x -1]`
`= (2 sin x xx cos x) /( 2 cos x xx cos x )`
` = (sin x) / (cos x) `
` = tan x = RHS`
\[\text{ Hence proved } .\]
APPEARS IN
संबंधित प्रश्न
Prove that: \[\frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x} = \tan x\]
Prove that: \[\frac{\cos 2 x}{1 + \sin 2 x} = \tan \left( \frac{\pi}{4} - x \right)\]
Prove that: \[\left( \sin 3x + \sin x \right) \sin x + \left( \cos 3x - \cos x \right) \cos x = 0\]
If \[\cos x = \frac{4}{5}\] and x is acute, find tan 2x
If \[\sin x = \frac{4}{5}\] and \[0 < x < \frac{\pi}{2}\]
, find the value of sin 4x.
If \[2 \tan \alpha = 3 \tan \beta,\] prove that \[\tan \left( \alpha - \beta \right) = \frac{\sin 2\beta}{5 - \cos 2\beta}\] .
If \[\cos x = \frac{\cos \alpha + \cos \beta}{1 + \cos \alpha \cos \beta}\] , prove that \[\tan\frac{x}{2} = \pm \tan\frac{\alpha}{2}\tan\frac{\beta}{2}\]
If \[a \cos2x + b \sin2x = c\] has α and β as its roots, then prove that
(ii) \[\tan\alpha \tan\beta = \frac{c - a}{c + a}\]
Prove that: \[4 \left( \cos^3 10 °+ \sin^3 20° \right) = 3 \left( \cos 10°+ \sin 2° \right)\]
\[\cot x + \cot\left( \frac{\pi}{3} + x \right) + \cot\left( \frac{2\pi}{3} + x \right) = 3 \cot 3x\]
Prove that: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]
Prove that: \[\cos 78° \cos 42° \cos 36° = \frac{1}{8}\]
Prove that: \[\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15} = \frac{1}{16}\]
Prove that: \[\cos 36° \cos 42° \cos 60° \cos 78° = \frac{1}{16}\]
Write the value of \[\cos^2 76° + \cos^2 16° - \cos 76° \cos 16°\]
Write the value of \[\cos\frac{\pi}{7} \cos\frac{2\pi}{7} \cos\frac{4\pi}{7} .\]
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 \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha - \cos \beta = b \text{ then } \tan \frac{\alpha - \beta}{2} =\]
The value of \[\tan x \sin \left( \frac{\pi}{2} + x \right) \cos \left( \frac{\pi}{2} - x \right)\]
If \[\tan \left( \pi/4 + x \right) + \tan \left( \pi/4 - x \right) = \lambda \sec 2x, \text{ then } \]
If \[\tan x = t\] then \[\tan 2x + \sec 2x =\]
The value of \[\tan x \tan \left( \frac{\pi}{3} - x \right) \tan \left( \frac{\pi}{3} + x \right)\] is
The value of \[\frac{\sin 5 \alpha - \sin 3\alpha}{\cos 5 \alpha + 2 \cos 4\alpha + \cos 3\alpha} =\]
If \[\text{ tan } x = \frac{a}{b}\], then \[b \cos 2x + a \sin 2x\]
The value of `cos^2 48^@ - sin^2 12^@` is ______.
If A = cos2θ + sin4θ for all values of θ, then prove that `3/4` ≤ A ≤ 1.
The greatest value of sin x cos x is ______.
If acos2θ + bsin2θ = c has α and β as its roots, then prove that tanα + tanβ = `(2b)/(a + c)`.
`["Hint: Use the identities" cos2theta = (1 - tan^2theta)/(1 + tan^2theta) "and" sin2theta = (2tantheta)/(1 + tan^2theta)]`.
Find the value of the expression `cos^4 pi/8 + cos^4 (3pi)/8 + cos^4 (5pi)/8 + cos^4 (7pi)/8`
[Hint: Simplify the expression to `2(cos^4 pi/8 + cos^4 (3pi)/8) = 2[(cos^2 pi/8 + cos^2 (3pi)/8)^2 - 2cos^2 pi/8 cos^2 (3pi)/8]`
If sinθ = `(-4)/5` and θ lies in the third quadrant then the value of `cos theta/2` is ______.
The value of cos248° – sin212° is ______.
[Hint: Use cos2A – sin2 B = cos(A + B) cos(A – B)]
If k = `sin(pi/18) sin((5pi)/18) sin((7pi)/18)`, then the numerical value of k is ______.
