Question

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HERE'S THE QUESTION FOR YOU ALL
PLEASE SOLVE IT.....


ITS OF FIIT JEE CLASS 10


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Answer 1

Consider : tan(α + β)

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{tan(A + B) = \dfrac{tanA + tanB}{1 - tanA.tanB}}}}$

$\implies \mathsf{tan(\alpha + \beta) = \dfrac{tan\alpha + tan\beta}{1 - tan\alpha.tan\beta}}$

$\textsf{Given :}\\\\\mathsf{\bigstar\;\;tan\alpha = \dfrac{m}{m + 1}}\\\\\\\mathsf{\bigstar\;\;tan\beta = \dfrac{1}{2m + 1}}$

$\mathsf{\implies tan(\alpha + \beta) = \dfrac{\bigg(\dfrac{m}{m + 1} + \dfrac{1}{2m + 1}\bigg)}{1 -\bigg(\dfrac{m}{m + 1}\bigg)\bigg(\dfrac{1}{2m + 1}\bigg)}}$

$\mathsf{\implies tan(\alpha + \beta) = \dfrac{\dfrac{m(2m + 1) + (m + 1)}{(m + 1)(2m + 1)}}{1 -\dfrac{m}{(m + 1)(2m + 1)}}}$

$\mathsf{\implies tan(\alpha + \beta) = \dfrac{\dfrac{m(2m + 1) + (m + 1)}{(m + 1)(2m + 1)}}{\dfrac{(m + 1)(2m + 1) - m}{(m + 1)(2m + 1)}}}$

$\mathsf{\implies tan(\alpha + \beta) = \dfrac{m(2m + 1) + (m + 1)}{(m + 1)(2m + 1) - m}}$

$\mathsf{\implies tan(\alpha + \beta) = \dfrac{2m^2 + m + m + 1}{2m^2 + m + 2m + 1 - m}}$

$\mathsf{\implies tan(\alpha + \beta) = \dfrac{2m^2 + 2m + 1}{2m^2 + 2m + 1}}$

$\mathsf{\implies tan(\alpha + \beta) = 1}$

$\mathsf{\implies (\alpha + \beta) = tan^{-1}(1)}$

$\mathsf{\implies (\alpha + \beta) = \dfrac{\pi}{4}}$

Answer 2

$\underline{\underline{\text{Using trigonometric laws :}}}$

$\text{Given that,}$

$\mathrm{tan\alpha=\frac{m}{m+1},\:tan\beta=\frac{1}{2m+1}}$

$\mathrm{Now,\:tan(\alpha+\beta)}$

$\mathrm{=\frac{tan\alpha+tan\beta}{1-tan\alpha\:tan\beta}}$

$\mathrm{=\frac{\frac{m}{m+1}+\frac{1}{2m+1}}{1-\frac{m}{m+1}\frac{1}{2m+1}}}$

$\mathrm{=\frac{m(2m+1)+1(m+1)}{(m+1)(2m+1)-m}}$

$\mathrm{=\frac{2m^{2}+m+m+1}{2m^{2}+m+2m+1-m}}$

$\mathrm{=\frac{2m^{2}+2m+1}{2m^{2}+2m+1}=1}$

$\implies \mathrm{tan(\alpha+\beta)=1}$

$\implies \mathrm{tan(\alpha+\beta)=tan\frac{\pi}{4}}$

$\therefore \boxed{\mathrm{\alpha+\beta=\frac{\pi}{4}}}$

$\underline{\text{Formula used :}}$

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

$\underline{\underline{\text{Another method :}}}$

$\mathrm{Here,\:tan\alpha=\frac{m}{m+1}}$

$\to \mathrm{(m+1)tan\alpha=m}$

$\to \mathrm{m\:tan\alpha+tan\alpha=m}$

$\to \mathrm{m(1-\:tan\alpha)=tan\alpha}$

$\to \mathrm{m =\frac{tan\alpha}{1-tan\alpha}}$

$\mathrm{Now,\:tan\beta=\frac{1}{2m+1}}$

$\to \mathrm{tan\beta=\frac{1}{\frac{2\:tan\alpha}{1-tan\alpha}+1}}$

$\to \mathrm{tan\beta=\frac{1-tan\alpha}{2\:tan\alpha+1-tan\alpha}}$

$\to \mathrm{tan\beta=\frac{1-tan\alpha}{1+tan\alpha}}$

$\to \mathrm{\frac{sin\beta}{cos\beta}=\frac{1-\frac{sin\alpha}{cos\alpha}}{1+\frac{sin\alpha}{cos\alpha}}}$

$\to \mathrm{\frac{sin\beta}{cos\beta}=\frac{cos\alpha-sin\alpha}{cos\alpha+sin\alpha}}$

$\to \mathrm{cos\alpha\:sin\beta+sin\alpha\:sin\beta}$
$\mathrm{=cos\alpha\:cos\beta-sin\alpha\:cos\beta}$

$\to \mathrm{sin\alpha\:cos\beta+cos\alpha\:sin\beta}$
$\mathrm{=cos\alpha\:cos\beta-sin\alpha\:sin\beta}$

$\to \mathrm{sin(\alpha+\beta)=cos(\alpha+\beta)}$

$\to \mathrm{tan(\alpha+\beta)=1}$

$\to \mathrm{\alpha+\beta=tan^{-1}(1)}$

$\implies \boxed{\mathrm{\alpha+\beta=\frac{\pi}{4}}}$