Question

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Answer 1

Given :-

$tan(A+B) = \dfrac{1}{3}$

$tan(A-B) = \dfrac{2}{5}$

To prove :-

$tan2A= \dfrac{11}{3}$

$tan2B = \dfrac{-1}{17}$

Proof :-

As they given tan(A+B) = 1/3 In R.H.S we cannot convert 1/3 in terms of tan as like tan(A-B) = 2/5 We cannot express the R.H.S in form of tan ratio .So, we shall solve into another method.

From A+B and A- B If we add these two equations we get

A+B + A-B = 2A

So,

tan(2A) can be written as tan(A+B+A-B)

By using formula,

$tan(A+B) = \dfrac{tanA+tanB}{1-tanAtanB}$

$tan (A+B+A-B) =\dfrac{tan(A+B)+ tan(A-B)}{1-tan(A+B)tan(A-B)}$

Substituting the values ,

• tan(A+B) = 1/3
• tan(A-B) = 2/5

$tan2A = \dfrac{\dfrac{1}{3}+\dfrac{2}{5} }{1-\bigg(\dfrac{1}{3}\bigg)\bigg(\dfrac{2}{5} \bigg) }$

$tan2A = \dfrac{\dfrac{5+6}{15} }{1-\bigg(\dfrac{2}{15}\bigg) }$

$tan2A = \dfrac{\dfrac{11}{15} }{\dfrac{15-2}{15} }$

$tan2A = \dfrac{\dfrac{11}{15} }{\dfrac{13}{15} }$

$tan2A = \dfrac{11}{13}$

Hence proved !

Now , for proving tan2B = -1/17

We shall subtract A+B and A-B in order to get 2B

A+B -(A-B) = A+B -A+B

= 2B

So,

A+B -(A-B) = 2B

So,

tan(2B) can be written as tan[A+B -(A-B)]

Solving it tan(A-B) formula that is

$tan(A-B) = \dfrac{tanA-tanB}{1+tanA tanB }$

$tan[A+B-(A-B)] = \dfrac{tan(A+B)-tan(A-B)}{1+tan(A+B)tan(A-B)}$

Substituting the values ,

• tan(A+B) = 1/3
• tan(A-B) = 2/5

$tan[A+B-(A-B)] = \dfrac{\dfrac{1}{3} -\dfrac{2}{5} }{1+\bigg(\dfrac{1}{3} \bigg)\bigg(\dfrac{2}{5}\bigg) }$

$tan(2B) =\dfrac{\dfrac{5-6}{15} }{1+\dfrac{2}{15} }$

$tan(2B) =\dfrac{\dfrac{-1}{15} }{\dfrac{15+2}{15} }$

$tan(2B) =\dfrac{\dfrac{-1}{15} }{\dfrac{17}{15} }$

$tan(2B) =\dfrac{-1}{17}$

Hence proved !

Used formulae:-

$tan(A+B) = \dfrac{tanA+tanB}{1-tanAtanB}$

$tan(A-B) = \dfrac{tanA-tanB}{1+tanA tanB }$

Know more formulae:-

Multiple angles :-

$sin2A = 2sinAcosA$

$sin2A = \dfrac{2tanA}{1 + tan {}^{2}A }$

$cos2A = cos {}^{2} A - sin { }^{2} A$

$cos2A= 2cos {}^{2} A - 1$

$cos2A = 1 - 2in {}^{2} A$

$cos2A = \dfrac{1 - tan {}^{2} A}{1 + tan {}^{2}A }$

$tan2A = \dfrac{2tanA}{1 - tan {}^{2}A }$

$cot2A = \dfrac{cot {}^{2} A - 1}{2cotA}$

$sin3A = 3sinA - 4sin {}^{3} A$

$cos3A = 4cos {}^{3} A - 3cosA$

$tan3A = \dfrac{3tanA - tan {}^{3}A }{1 - 3tan {}^{2} A}$

Compound angles:-

$sin(A+B)= sinAcosB + sinBcosA$

$sin(A-B) = sinAcosB- sinBcosA$

$cos(A+B) = cosAcosB - sinAsinB$

$cos(A-B) = cosAcosB + sinAsinB$

$tan(A+B) = \dfrac{tanA+tanB}{1-tanAtanB}$

$tan(A-B) = \dfrac{tanA-tanB}{1+tanAtanB}$

$cot(A+B) = \dfrac{cotB cotA -1}{cotB + cotA}$

$cot(A-B) = \dfrac{cotB cotA + 1}{cotB-cotA}$