Advertisements
Advertisements
प्रश्न
If \[\tan x = t\] then \[\tan 2x + \sec 2x =\]
विकल्प
- \[\frac{1 + t}{1 - t}\]
- \[\frac{1 - t}{1 + t}\]
- \[\frac{2t}{1 - t}\]
- \[\frac{2t}{1 + t}\]
Advertisements
उत्तर
\[\frac{1 + t}{1 - t}\]
\[\tan 2x + \sec 2x = \frac{2 \tan x}{1 - \tan^2 x} + \frac{1 + \tan^2 x}{1 - \tan^2 x}\]
\[ = \frac{2\tan x + 1 + \tan^2 x}{1 - \tan^2 x}\]
\[ = \frac{\left( 1 + \tan x \right)^2}{1 - \tan^2 x}\]
\[ = \frac{\left( 1 + \tan x \right)\left( 1 + \tan x \right)}{\left( 1 + \tan x \right)\left( 1 - \tan x \right)}\]
\[ = \frac{1 + \tan x}{1 - \tan x}\]
\[ = \frac{1 + t}{1 - t} \left[ \tan x = t \left( \text{ given } \right) \right]\]
APPEARS IN
संबंधित प्रश्न
Prove that: \[\sqrt{2 + \sqrt{2 + 2 \cos 4x}} = 2 \text{ cos } x\]
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: \[\frac{\cos 2 x}{1 + \sin 2 x} = \tan \left( \frac{\pi}{4} - x \right)\]
Prove that: \[\left( \cos \alpha + \cos \beta^2 \right) + \left( \sin \alpha + \sin \beta \right)^2 = 4 \cos^2 \left( \frac{\alpha - \beta}{2} \right)\]
Prove that: \[\cos^3 2x + 3 \cos 2x = 4\left( \cos^6 x - \sin^6 x \right)\]
Prove that: \[\cos^2 \left( \frac{\pi}{4} - x \right) - \sin^2 \left( \frac{\pi}{4} - x \right) = \sin 2x\]
Prove that: \[\cos^6 A - \sin^6 A = \cos 2A\left( 1 - \frac{1}{4} \sin^2 2A \right)\]
Prove that: \[\cos 4x - \cos 4\alpha = 8 \left( \cos x - \cos \alpha \right) \left( \cos x + \cos \alpha \right) \left( \cos x - \sin \alpha \right) \left( \cos x + \sin \alpha \right)\]
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 \[\tan A = \frac{1}{7}\] 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 \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha + \cos \beta = b\] , prove that
(ii) \[\cos \left( \alpha - \beta \right) = \frac{a^2 + b^2 - 2}{2}\]
If \[\sin \alpha = \frac{4}{5} \text{ and } \cos \beta = \frac{5}{13}\] , prove that \[\cos\frac{\alpha - \beta}{2} = \frac{8}{\sqrt{65}}\]
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: \[4 \left( \cos^3 10 °+ \sin^3 20° \right) = 3 \left( \cos 10°+ \sin 2° \right)\]
\[\cot x + \cot\left( \frac{\pi}{3} + x \right) + \cot\left( \frac{2\pi}{3} + x \right) = 3 \cot 3x\]
Prove that \[\left| \sin x \sin \left( \frac{\pi}{3} - x \right) \sin \left( \frac{\pi}{3} + x \right) \right| \leq \frac{1}{4}\] for all values of x
Prove that: \[\sin^2 \frac{2\pi}{5} - \sin^{2 -} \frac{\pi}{3} = \frac{\sqrt{5} - 1}{8}\]
If \[\frac{\pi}{4} < x < \frac{\pi}{2}\], then write the value of \[\sqrt{1 - \sin 2x}\] .
The value of \[\cos \frac{\pi}{65} \cos \frac{2\pi}{65} \cos \frac{4\pi}{65} \cos \frac{8\pi}{65} \cos \frac{16\pi}{65} \cos \frac{32\pi}{65}\] is
The value of \[2 \tan \frac{\pi}{10} + 3 \sec \frac{\pi}{10} - 4 \cos \frac{\pi}{10}\] is
The value of \[2 \sin^2 B + 4 \cos \left( A + B \right) \sin A \sin B + \cos 2 \left( A + B \right)\] is
The value of \[\frac{2\left( \sin 2x + 2 \cos^2 x - 1 \right)}{\cos x - \sin x - \cos 3x + \sin 3x}\] is
If \[\tan \frac{x}{2} = \frac{\sqrt{1 - e}}{1 + e} \tan \frac{\alpha}{2}\] , then \[\cos \alpha =\]
If \[\left( 2^n + 1 \right) x = \pi,\] then \[2^n \cos x \cos 2x \cos 2^2 x . . . \cos 2^{n - 1} x = 1\]
If \[\text{ tan } x = \frac{a}{b}\], then \[b \cos 2x + a \sin 2x\]
The greatest value of sin x cos x is ______.
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]`
The value of `sin pi/10 sin (13pi)/10` is ______.
`["Hint: Use" sin18^circ = (sqrt5 - 1)/4 "and" cos36^circ = (sqrt5 + 1)/4]`
The value of `sin pi/18 + sin pi/9 + sin (2pi)/9 + sin (5pi)/18` is given by ______.
The value of cos248° – sin212° is ______.
[Hint: Use cos2A – sin2 B = cos(A + B) cos(A – B)]
