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Prove that

Question:

Prove that $\frac{\tan 69^{\circ}+\tan 66^{\circ}}{1-\tan 69^{\circ} \tan 66^{\circ}}=-1$.

Solution:

$\mathrm{LHS}=\frac{\tan 69^{\circ}+\tan 66^{\circ}}{1-\tan 69^{\circ} \tan 66^{\circ}}$

$=\tan \left(69^{\circ}+66^{\circ}\right) \quad\left[\mathrm{U} \operatorname{sing}\right.$ the formula $\left.\frac{\tan A+\tan B}{1-\tan A \tan B}=\tan (A+B)\right]$

$=\tan 135^{\circ}$

$=\tan \left(180^{\circ}-45^{\circ}\right)$

$=-\tan 45^{\circ} \quad[\tan (180-A)=-\tan A]$

$=-1$

= RHS

Hence proved.

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