If tan 80^{circ} = alpha and tan 47^{circ} = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $	an 80^{\circ} = \alpha$. Since $	an 80^{\circ} = \cot 10^{\circ}$,we have $\cot 10^{\circ} = \alpha$,which implies $	an 10^{\circ} = \f

Vedclass · Accepted Answer

$\frac{\alpha \beta - 1}{\alpha + \beta}$

Vedclass · Answer

(C) Given $	an 80^{\circ} = \alpha$. Since $	an 80^{\circ} = \cot 10^{\circ}$,we have $\cot 10^{\circ} = \alpha$,which implies $	an 10^{\circ} = \frac{1}{\alpha}$.Given $	an 47^{\circ} = \beta$.We need to find $	an 37^{\circ}$. We can write $37^{\circ} = 47^{\circ} - 10^{\circ}$.Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$:$	an 37^{\circ} = 	an(47^{\circ} - 10^{\circ}) = \frac{	an 47^{\circ} - 	an 10^{\circ}}{1 + 	an 47^{\circ} 	an 10^{\circ}}$Substituting the values:$= \frac{\beta - \frac{1}{\alpha}}{1 + \beta \cdot \frac{1}{\alpha}} = \frac{\frac{\alpha \beta - 1}{\alpha}}{\frac{\alpha + \beta}{\alpha}} = \frac{\alpha \beta - 1}{\alpha + \beta}$

If $\tan 80^{\circ} = \alpha$ and $\tan 47^{\circ} = \beta$,then $\tan 37^{\circ}$ is equal to -

Similar Questions

If $\cos (A - B) = \frac{3}{5}$ and $\tan A \tan B = 2$,then

If $\cos (\alpha+\beta)=\frac{4}{5}$,$\sin (\alpha-\beta)=\frac{5}{13}$ and $\alpha, \beta$ are between $0$ and $\frac{\pi}{4}$,then $\tan 2 \alpha$ is equal to

Prove that $\frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} = -\frac{\sin 2x}{\cos 10x}$

If $\frac{\pi}{2} < \alpha < \pi$ and $\pi < \beta < \frac{3\pi}{2}$,with $\sin \alpha = \frac{15}{17}$ and $\tan \beta = \frac{12}{5}$,then the value of $\sin(\beta - \alpha)$ is (in $/221$)

$\sin \left(\frac{\pi}{3}+x\right)-\cos \left(\frac{\pi}{6}+x\right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module