Math, asked by toni8190, 4 days ago

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

Answers

Answered by Anonymous
19

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}

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