Question

10. Prove that:
tanA+tanb÷tanA-tanB=sin(A+B)÷sin(A-B)​

Answer 1

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Answer 2

$\huge\mathfrak{Answer:}$

$\large\underline{\sf{\blue{Given:}}}$

• We have been given a trigonometric equation
• Containing some Trigonometric function

$\large\underline{\sf{\blue{To \: Prove:}}}$

• $\sf{\dfrac{tanA + tanB}{tanA - tanB} = \dfrac{sin(A+B)}{sin(A-B)}}$

$\large\underline{\sf{\blue{Proof:}}}$

We have been given a Trigonometric equation :

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$\large\underline{\mathfrak{\red{Taking \: Left \: Hand \: Side:}}}$

$\implies \sf{\dfrac{tanA + tanB}{tanA - tanB}}$

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Converting the tangent component in terms of sine and cosine component

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$\implies \sf{\dfrac{\frac{sinA}{cosA}+ \frac{sinB}{cosB}}{\frac{sinA}{cosA} - \frac{sinB}{cosB}}}$

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$\implies \sf{\dfrac{\frac{sinA \: cosB + cosA \: sinB}{cosA \: cosB}}{\frac{sinA \: cosB - cosA \: sinB}{cosA \: cosB}}}$

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On Reciprocaling the denominator of fraction we get

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$\sf{\dfrac{sinA \: cosB + cosA \: sinB}{cosA \: cosB} \times \dfrac{cosA \: cosB}{sinA \: cosB - cosA \: sinB}}$

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$\sf{\dfrac{sinA \: cosB + cosA \: sinB}{\cancel{cosA \: cosB} } \times \dfrac{\cancel{cosA \: cosB} }{sinA \: cosB - cosA \: sinB }}$

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$\sf{\dfrac{sinA \: cosB + cosA \: sinB}{sinA \: cosB - cosA \: sinB}}$

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Using following trigonometric Identity

$\boxed{\sf{\green{sin(A+B) = sinA \: cosB + cosA \: sinB} }}$

$\boxed{\sf{\green{sin(A-B) = sinA \: cosB - cosA \: sinB}} }$

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$\implies \sf{\dfrac{sin(A+B) }{sin(A-B) } }$

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$\implies \sf{R. H. S}$

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Hence it is Proved that :

$\boxed{\sf{\dfrac{tanA + tanB}{tanA - tanB} = \dfrac{sin(A+B)}{sin(A-B)}}}$

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$\large\underline{\mathfrak{\red{Some \: Trigonometric \: Identities :}}}$

$\sf{cos(A+B) = cosA \: cosB - sinA \: sinB}$

$\sf{cos(A-B) = cosA \: cosB + sinA \: sinB}$

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

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

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