જો $3 \sin (\alpha-\beta)=5 \cos (\alpha+\beta)$ અને $\alpha+\beta \neq \frac{\pi}{2}$ હોય,તો $\frac{\tan \left(\frac{\pi}{4}-\alpha\right)}{\tan \left(\frac{\pi}{4}-\beta\right)}=$

  • A
    $0$
  • B
    $-4$
  • C
    $-\frac{1}{4}$
  • D
    $\frac{1}{2}$

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Similar Questions

જો $\cos \theta + \sin \theta = \sqrt{2} \cos \theta$ અને $0 < \theta < \frac{\pi}{2}$ હોય,તો $\sec 2 \theta + \tan 2 \theta = $

જો $x = a \cos^3 \theta$ અને $y = b \sin^3 \theta$ હોય,તો:

$\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{2 \pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{4 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{6 \pi}{8}+\cos ^4 \frac{7 \pi}{8}+\cos ^4 \frac{8 \pi}{8}$ ની કિંમત શોધો.

પદાવલિનું મૂલ્ય શોધો: $\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{2 \pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{4 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{6 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)$

જો $\sin \theta + \cos \theta = x$ હોય,તો ${\sin ^6}\theta + {\cos ^6}\theta = \frac{1}{4}[4 - 3{({x^2} - 1)^2}]$ એ . . . માટે સાચું છે.

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