Advertisements
Advertisements
प्रश्न
If \[\tan \frac{x}{2} = \frac{\sqrt{1 - e}}{1 + e} \tan \frac{\alpha}{2}\] , then \[\cos \alpha =\]
विकल्प
\[1 - e \cos \left( \cos x + e \right)\]
\[\frac{1 + e \cos x}{\cos x - e}\]
\[\frac{1 - e \cos x}{\cos x - e}\]
\[\frac{\cos x - e}{1 - e \cos x}\]
Advertisements
उत्तर
\[\frac{\cos x - e}{1 - e \cos x}\]
\[\text { Given } : \tan\frac{x}{2} = \sqrt{\frac{1 - e}{1 + e}}\tan\frac{\alpha}{2}\]
\[ \Rightarrow \frac{\tan\frac{x}{2}}{\tan\frac{\alpha}{2}} = \sqrt{\frac{1 - e}{1 + e}}\]
\[\text{ Squaring both sides, we get, } \]
\[\frac{\tan^2 \frac{x}{2}}{\tan^2 \frac{\alpha}{2}} = \frac{1 - e}{1 + e}\]
\[ \Rightarrow \tan^2 \frac{\alpha}{2}\left( 1 - e \right) = \tan^2 \frac{x}{2}\left( 1 + e \right)\]
\[\Rightarrow \frac{\sin^2 \frac{\alpha}{2}}{\cos^2 \frac{\alpha}{2}}\left( 1 - e \right) = \frac{\sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2}}\left( 1 + e \right)\]
\[ \Rightarrow \frac{\frac{1}{2}\left( 1 - cos\alpha \right)}{\frac{1}{2}\left( 1 + cos\alpha \right)}\left( 1 - e \right) = \frac{\frac{1}{2}\left( 1 - \text{ cos } x \right)}{\frac{1}{2}\left( 1 + \text{ cos } x \right)}\left( 1 + e \right)\]
\[ \Rightarrow \left( 1 - cos\alpha \right)\left( 1 + \text{ cos } x \right)\left( 1 - e \right) = \left( 1 + cos\alpha \right)\left( 1 - \text{ cos } x \right)\left( 1 + e \right)\]
\[ \Rightarrow \left( 1 + \text{ cos } x \right)\left( 1 - e \right) - cos\alpha\left( 1 + \text{ cos } x \right)\left( 1 - e \right) = \left( 1 - \text{ cos } x \right)\left( 1 + e \right) + cos\alpha\left( 1 - \text{ cos } x \right)\left( 1 + e \right)\]
\[ \Rightarrow cos\alpha\left\{ \left( 1 + \text{ cos } x \right)\left( 1 - e \right) + \left( 1 - \text{ cos } x \right)\left( 1 + e \right) \right\} = \left( 1 + \text{ cos } x \right)\left( 1 - e \right) - \left( 1 - \text{ cos } x \right)\left( 1 + e \right)\]
\[ \Rightarrow cos\alpha = \frac{2\text{ cos } x - 2e}{2 - 2ecosx} = \frac{\text{ cos } x - e}{1 - ecosx}\]
APPEARS IN
संबंधित प्रश्न
Prove that: \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} = \tan x\]
Prove that: \[\frac{1 - \cos 2x + \sin 2x}{1 + \cos 2x + \sin 2x} = \tan x\]
Prove that: \[\frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x} = \tan x\]
Prove that: \[\left( \cos \alpha + \cos \beta^2 \right) + \left( \sin \alpha + \sin \beta \right)^2 = 4 \cos^2 \left( \frac{\alpha - \beta}{2} \right)\]
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 \[\tan A = \frac{1}{7}\] and \[\tan B = \frac{1}{3}\] , show that cos 2A = sin 4B
Prove that: \[\cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos \frac{8\pi}{15} \cos \frac{16\pi}{15} = \frac{1}{16}\]
If \[2 \tan \alpha = 3 \tan \beta,\] prove that \[\tan \left( \alpha - \beta \right) = \frac{\sin 2\beta}{5 - \cos 2\beta}\] .
If \[2 \tan\frac{\alpha}{2} = \tan\frac{\beta}{2}\] , prove that \[\cos \alpha = \frac{3 + 5 \cos \beta}{5 + 3 \cos \beta}\]
If \[a \cos2x + b \sin2x = c\] has α and β as its roots, then prove that
(i) \[\tan\alpha + \tan\beta = \frac{2b}{a + c}\]
Prove that: \[\cos^3 x \sin 3x + \sin^3 x \cos 3x = \frac{3}{4} \sin 4x\]
Prove that: \[\sin^2 \frac{2\pi}{5} - \sin^{2 -} \frac{\pi}{3} = \frac{\sqrt{5} - 1}{8}\]
Prove that: \[\sin^2 42° - \cos^2 78 = \frac{\sqrt{5} + 1}{8}\]
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: \[\cos 6° \cos 42° \cos 66° \cos 78° = \frac{1}{16}\]
Prove that: \[\cos 36° \cos 42° \cos 60° \cos 78° = \frac{1}{16}\]
If \[\tan\frac{x}{2} = \frac{m}{n}\] , then write the value of m sin x + n cos x.
If \[\frac{\pi}{2} < x < \pi,\] the write the value of \[\sqrt{2 + \sqrt{2 + 2 \cos 2x}}\] in the simplest form.
If \[\pi < x < \frac{3\pi}{2}\], then write the value of \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}\] .
In a right angled triangle ABC, write the value of sin2 A + Sin2 B + Sin2 C.
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
For all real values of x, \[\cot x - 2 \cot 2x\] is equal to
If \[\cos x = \frac{1}{2} \left( a + \frac{1}{a} \right),\] and \[\cos 3 x = \lambda \left( a^3 + \frac{1}{a^3} \right)\] then \[\lambda =\]
The value of \[\tan x \sin \left( \frac{\pi}{2} + x \right) \cos \left( \frac{\pi}{2} - x \right)\]
If \[5 \sin \alpha = 3 \sin \left( \alpha + 2 \beta \right) \neq 0\] , then \[\tan \left( \alpha + \beta \right)\] is equal to
The value of \[\cos^4 x + \sin^4 x - 6 \cos^2 x \sin^2 x\] is
The value of \[\cos \left( 36° - A \right) \cos \left( 36° + A \right) + \cos \left( 54° - A \right) \cos \left( 54° + A \right)\] 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 \[\text{ tan } x = \frac{a}{b}\], then \[b \cos 2x + a \sin 2x\]
If \[\tan\alpha = \frac{1}{7}, \tan\beta = \frac{1}{3}\], then
\[\cos2\alpha\] is equal to
The value of `cos pi/5 cos (2pi)/5 cos (4pi)/5 cos (8pi)/5` is ______.
The value of cos12° + cos84° + cos156° + cos132° is ______.
The value of cos248° – sin212° is ______.
[Hint: Use cos2A – sin2 B = cos(A + B) cos(A – B)]
