Advertisements
Advertisements
प्रश्न
Prove that: \[\frac{\cos 2 x}{1 + \sin 2 x} = \tan \left( \frac{\pi}{4} - x \right)\]
Advertisements
उत्तर
\[LHS = \frac{\cos2x}{1 + \sin2x}\]
\[= \frac{\cos^2 x - \sin^2 x}{\sin^2 x + \cos^2 x + 2\text{ sin } x \times \text{ cos } x} \left[ \because \cos2x = \cos^2 x - \sin^2 x \text{ and } \sin^2 x + \cos^2 x = 1 \right]\]
\[ = \frac{\left( \text{ cos } x - \text{ sin } x \right)\left( \text{ cos } x + \text{ sin } x \right)}{\left( \text{ cos } x + \text{ sin } x \right)^2} \left[ \because a^2 - b^2 = \left( a + b \right)\left( a - b \right) \right]\]
\[ = \frac{\text{ cos } x - \text{ sin } x}{\text{ cos } x + \text{ sin } x}\]
On dividing the numerator and denominator by cos x, we get
\[= \frac{1 - \text{ tan } x}{1 + \text{ tan } x}\]
\[ = \tan\left( \frac{\pi}{4} - x \right) = RHS\]
\[\text{ Hence proved } .\]
APPEARS IN
संबंधित प्रश्न
Prove that: \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} = \tan x\]
Prove that: \[\frac{\sin 2x}{1 + \cos 2x} = \tan x\]
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: \[\frac{\cos x}{1 - \sin x} = \tan \left( \frac{\pi}{4} + \frac{x}{2} \right)\]
Prove that: \[\cos^2 \frac{\pi}{8} + \cos^2 \frac{3\pi}{8} + \cos^2 \frac{5\pi}{8} + \cos^2 \frac{7\pi}{8} = 2\]
Prove that: \[\sin^2 \frac{\pi}{8} + \sin^2 \frac{3\pi}{8} + \sin^2 \frac{5\pi}{8} + \sin^2 \frac{7\pi}{8} = 2\]
Prove that: \[1 + \cos^2 2x = 2 \left( \cos^4 x + \sin^4 x \right)\]
Show that: \[3 \left( \sin x - \cos x \right)^4 + 6 \left( \sin x + \cos \right)^2 + 4 \left( \sin^6 x + \cos^6 x \right) = 13\]
Prove that: \[\cos^6 A - \sin^6 A = \cos 2A\left( 1 - \frac{1}{4} \sin^2 2A \right)\]
If 0 ≤ x ≤ π and x lies in the IInd quadrant such that \[\sin x = \frac{1}{4}\]. Find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2} \text{ and } \tan\frac{x}{2}\]
If \[\cos x = \frac{4}{5}\] and x is acute, find tan 2x
If \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha + \cos \beta = b\] , prove that
(i)\[\sin \left( \alpha + \beta \right) = \frac{2ab}{a^2 + b^2}\]
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 \[\cos \alpha + \cos \beta = \frac{1}{3}\] and sin \[\sin\alpha + \sin \beta = \frac{1}{4}\] , prove that \[\cos\frac{\alpha - \beta}{2} = \pm \frac{5}{24}\]
If \[\sin \alpha = \frac{4}{5} \text{ and } \cos \beta = \frac{5}{13}\] , prove that \[\cos\frac{\alpha - \beta}{2} = \frac{8}{\sqrt{65}}\]
Prove that: \[4 \left( \cos^3 10 °+ \sin^3 20° \right) = 3 \left( \cos 10°+ \sin 2° \right)\]
Prove that: \[\cos^3 x \sin 3x + \sin^3 x \cos 3x = \frac{3}{4} \sin 4x\]
Prove that: \[\sin^2 42° - \cos^2 78 = \frac{\sqrt{5} + 1}{8}\]
Prove that: \[\cos 6° \cos 42° \cos 66° \cos 78° = \frac{1}{16}\]
If \[\cos 4x = 1 + k \sin^2 x \cos^2 x\] , then write the value of k.
If \[\text{ sin } x + \text{ cos } x = a\], find the value of \[\left|\text { sin } x - \text{ cos } x \right|\] .
If \[2 \tan \alpha = 3 \tan \beta, \text{ then } \tan \left( \alpha - \beta \right) =\]
If \[\tan \alpha = \frac{1 - \cos \beta}{\sin \beta}\] , then
\[2 \text{ cos } x - \ cos 3x - \cos 5x - 16 \cos^3 x \sin^2 x\]
If \[\tan x = t\] then \[\tan 2x + \sec 2x =\]
The value of \[\cos^4 x + \sin^4 x - 6 \cos^2 x \sin^2 x\] 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 \[b \cos 2x + a \sin 2x\]
Prove that sin 4A = 4sinA cos3A – 4 cosA sin3A
If tan(A + B) = p, tan(A – B) = q, then show that tan 2A = `(p + q)/(1 - pq)`
If k = `sin(pi/18) sin((5pi)/18) sin((7pi)/18)`, then the numerical value of k is ______.
