If tan 80^{circ} = a and tan 47^{circ} = b,th | vedclass

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(C) Given: $	an 80^{\circ} = a$ and $	an 47^{\circ} = b$.We know that $	an 80^{\circ} = \cot 10^{\circ} = \frac{1}{	an 10^{\circ}}$,therefore $	a

Vedclass · Accepted Answer

$\frac{ab - 1}{a + b}$

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(C) Given: $	an 80^{\circ} = a$ and $	an 47^{\circ} = b$.We know that $	an 80^{\circ} = \cot 10^{\circ} = \frac{1}{	an 10^{\circ}}$,therefore $	an 10^{\circ} = \frac{1}{a}$.We need to find $	an 37^{\circ}$.We can write $	an 37^{\circ} = 	an(47^{\circ} - 10^{\circ})$.Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we get:$	an 37^{\circ} = \frac{	an 47^{\circ} - 	an 10^{\circ}}{1 + 	an 47^{\circ} 	an 10^{\circ}}$.Substituting the values,we have:$	an 37^{\circ} = \frac{b - \frac{1}{a}}{1 + b \cdot \frac{1}{a}}$.Multiplying the numerator and denominator by $a$,we get:$	an 37^{\circ} = \frac{ab - 1}{a + b}$.

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

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