If tan A-56- tan B-111- prove that - A-B-4

We have, tan A=56, tan B=111 We know that tan(A+B)=tan A+tan B1 tan A tan B Therefore, tan(A+B)=56+1111 56×111 tan(A+B)=55+6661 566 tan(A+ ...

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Standard IX

Mathematics

Use of Trigonometric Table

If tan A=56...

Question

If tan $A = \frac{5}{6},$ tan $B = \frac{1}{11},$ prove that : $A + B = \frac{π}{4}$

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Solution

We have,
$tan A = \frac{5}{6}, tan B = \frac{1}{11}$

We know that
$tan (A + B) = \frac{tan A + tan B}{1 - tan A tan B}$

Therefore,
$tan (A + B) = \frac{\frac{5}{6} + \frac{1}{11}}{1 - \frac{5}{6} \times \frac{1}{11}}$
$tan (A + B) = \frac{\frac{55 + 6}{66}}{1 - \frac{5}{66}}$
$tan (A + B) = \frac{\frac{61}{66}}{\frac{66 - 5}{66}}$
$tan (A + B) = 1$
$tan (A + B) = tan \frac{π}{4}$
$A + B = \frac{π}{4}$

Hence, proved.

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If tan A=56, tan B=111, prove that : A+B=π4

We have,tanA=56,tanB=111We know thattan(A+B)=tanA+tanB1−tanAtanBTherefore,tan(A+B)=56+1111−56×111tan(A+B)=55+6661−566tan(A+B)=616666−566tan(A+B)=1tan(A+B)=tanπ4A+B=π4Hence, proved.

If tan $A = \frac{5}{6},$ tan $B = \frac{1}{11},$ prove that : $A + B = \frac{π}{4}$