Question

cos alpha + cos beta equal to b and sin alpha + sin beta equal to a then prove that cos (alpha + beta) equal to b square minus A square upon B square + A square

Answer 1

$\textbf{Given:}$

$a=sin\,\alpha+sin\,\beta$

$b=cos\,\alpha+cos\,\beta$

$\textbf{To prove:}$

$cos(\alpha+\beta)=\dfrac{b^2-a^2}{b^2+a^2}$

$\textbf{Solution:}$

$\text{Consider,}$

$\dfrac{b^2-a^2}{b^2+a^2}$

$=\dfrac{(cos\,\alpha+cos\,\beta)^2-(sin\,\alpha+sin\,\beta)^2}{(cos\,\alpha+cos\,\beta)^2+(sin\,\alpha+sin\,\beta)^2}$

$=\dfrac{cos^2\alpha+cos^2\beta+2\,cos\,\alpha\,cos\,\beta-sin^2\alpha-sin^2\beta-2\,sin\,\alpha\,sin\,\beta}{cos^2\alpha+cos^2\beta+2\,cos\,\alpha\,cos\,\beta+sin^2\alpha+sin^2\beta+2\,sin\,\alpha\,sin\,\beta}$

$=\dfrac{(cos^2\alpha--sin^2\alpha)+(cos^2\beta-sin^2\beta)+2(cos\,\alpha\,cos\,\beta-\sin\,\alpha\,sin\,\beta)}{(cos^2\alpha+sin^2\alpha)+(cos^2\beta+sin^2\beta)+2(cos\,\alpha\,cos\,\beta+\sin\,\alpha\,sin\,\beta)}$

$\text{Using the following identities,}$

$\boxed{\bf\,cos2A=cos^2A-sin^2A}$

$\boxed{\bf\,cos^2A+sin^2A=1}$

$\boxed{\bf\,cos(A+B)=cosA\,cosB-sinA\,sinB}$

$\boxed{\bf\,cos(A-B)=cosA\,cosB+sinA\,sinB}$

$=\dfrac{cos\,2\alpha+cos\,2\beta+2\,cos(\alpha+\beta)}{1+1+2\,cos(\alpha-\beta)}$

$=\dfrac{(cos\,2\alpha+cos\,2\beta)+2\,cos(\alpha+\beta)}{2+2\,cos(\alpha-\beta)}$

$\text{Using the formula,}$

$\boxed{\bf\,cos\,C+cos\,D=2\,cos(\frac{C+D}{2})\,cos(\frac{C+D}{2})}$

$=\dfrac{2\,cos(\alpha+\beta)\,cos(\alpha-\beta)+2\,cos(\alpha+\beta)}{2+2\,cos(\alpha-\beta)}$

$=\dfrac{cos(\alpha+\beta)(2\,cos(\alpha-\beta)+2)}{2+2\,cos(\alpha-\beta)}$

$=cos(\alpha+\beta)$

$\implies\bf\,cos(\alpha+\beta)=\dfrac{b^2-a^2}{b^2+a^2}$

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