Question

Prove that tan(π/4+a)-tan(π/4-a)=2tan2a

Answer 1

Given to prove that :-

$\sf tan \bigg( \dfrac{ \pi}{4} - A \bigg) - tan \bigg( \dfrac{ \pi}{4} - A \bigg) = 2tanA$

Formulae used :-

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

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

$\sf \pi = 180 ^{ \circ}$

tan 45° = 1

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

Algebraic identities:-

(a +b)(a-b) = a² - b²

(a+b)² = a^2+2ab+b^2

(a-b)^2 = a^2 -2ab +b2

SOLUTION:-

By using above formulae We shall solve this question

Take L.H.S

$\sf tan \bigg( \dfrac{ \pi}{4} + A \bigg) - tan \bigg( \dfrac{ \pi}{4} - A \bigg)$

$\sf tan \bigg( \dfrac{ 180 ^{ \circ} }{4} + A \bigg) - tan \bigg( \dfrac{ 180 ^{ \circ}}{4} - A \bigg)$

$\sf tan(45 ^{ \circ} + A) - tan(45 ^{ \circ} - A)$

Applying the formulae tan(A+B) to the first term and tan(A-B) to the second term

$\sf = \bigg(\dfrac{tan45 ^{ \circ} + tanA}{1 - tan45 ^{ \circ} tanA} \bigg) - \bigg(\dfrac{tan45 ^{ \circ} - tanA}{1 + tan45 ^{ \circ} tanA} \bigg) -$

$\sf = \bigg(\dfrac{1 + tanA}{1 - tanA} \bigg) - \bigg(\dfrac{1 - tanA}{1 + tanA} \bigg)$

$\sf \dfrac{(1 + tanA) {}^{2} - (1 - tanA) {}^{2} }{(1 - tanA)(1 + tanA)}$

Expanding the numerator by (a+b)^2 and (a-b)^2

Simplifying the denominator by (a+b)(a-b) = a²-b²

$\sf = \dfrac{(1)^2+(tanA)^2+2(1)(tanA) - [(1)^2+(tanA)^2-2(1)(tanA)]}{1-tan^2A}$

$\sf =\dfrac{1+tan^2A+2tanA-[ 1+tan^2A-2tanA]}{1-tan^2A}$

$\sf =\dfrac{\not 1+\not tan^2A+2tanA\not -1\not -tan^2A+2tanA}{1-tan^2A}$

$\sf = \dfrac{4tanA}{1 - tan {}^{2}A }$

Take out the common 2

$\sf= 2 \bigg( \dfrac{2tanA}{1 - tan {}^{2} A} \bigg)$

$\sf = 2(tan2A)$

$\sf = 2tan2A$

Hence proved!

Know more formulae:-

❒Multiple angles :-

$\sf sin2A = 2sinAcosA$

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

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

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

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

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

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

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

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

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

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

❒Compound angles

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

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

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

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

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

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

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

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