If tan A - a - a -1 and tan B -1 -2 a -1- then the value of A - B is-

Tan A = a / (a + 1) tan B = 1 / (2 a + 1) Cot A = 1 + 1/a Cot B = 2a +1 Tan (A+B) = (Tan A + Tan B) / [1 Tan A tan B] ...

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Standard X

Mathematics

Range of Trigonometric Ratios from 0 to 90 Degrees

If tan A = a ...

Question

If $tan A = \frac{a}{(a + 1)}$ and $tan B = \frac{1}{(2 a + 1)}$ , then the value of $A + B$ is?

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Solution

$tan A = \frac{a}{(a + 1)}$
$cot A = \frac{a + 1}{(a)} = 1 + \frac{1}{a}$ [ $∵ cot A = \frac{1}{tan A}$ ]
and, $tan B = \frac{1}{(2 a + 1)}$
$cot B = 2 a + 1$

$tan (A + B)$
$= \frac{(tan A + tan B)}{[1 - tan A tan B]}$
$= \frac{(cot A + cot B)}{[cot A cot B - 1]}$ [On dividing numerator and denominator by $tan A tan B$ ]
$= \frac{(1 + \frac{1}{a} + 2 a + 1)}{[(1 + \frac{1}{a}) \times (2 a + 1) - 1]}$
$= \frac{\frac{2 a^{2} + 2 a + 1}{a}}{\frac{2 a^{2} + 2 a + 1}{a}}$
$= 1$
$tan (A + B) = 1$
$∴ A + B = \frac{π}{4}$ or $\frac{5 π}{4}$ or $n π + \frac{π}{4}$

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If tanA=a(a+1) and tanB=1(2a+1), then the value of A+B is?

tanA=a(a+1)cotA=a+1(a)=1+1a [∵cotA=1tanA]and, tanB=1(2a+1)cotB=2a+1 tan(A+B) =(tanA+tanB)[1−tanAtanB]=(cotA+cotB)[cotAcotB−1] [On dividing numerator and denominator by tanAtanB] =(1+1a+2a+1)[(1+1a)×(2a+1)−1]=2a2+2a+1a2a2+2a+1a=1tan(A+B)=1∴A+B=π4 or 5π4 or nπ+π4

If $tan A = \frac{a}{(a + 1)}$ and $tan B = \frac{1}{(2 a + 1)}$ , then the value of $A + B$ is?