Math, asked by taneesha76, 7 hours ago

if a,b,c are angles in a triangle, then prove that cosA+cosB-cosC = -1+4 cos A/2 cos B/2 sin C/2​

Answers

Answered by MaheswariS
2

\underline{\textbf{Given:}}

\textsf{A,B,C are angles of a triangle}

\underline{\textbf{To prove:}}

\mathsf{cosA+cosB-cosC=-1+4\,cos\dfrac{A}{2}\,cos\dfrac{B}{2}\,sin\dfrac{C}{2}}

\underline{\textbf{Solution:}}

\underline{\textbf{Formula used:}}

\mathsf{1.\;cosA=1-2\,sin^2\dfrac{A}{2}}

\mathsf{2.\;cosC+cosD=2\,cos\left(\dfrac{C+D}{2}\right)\;cos\left(\dfrac{C-D}{2}\right)}

\mathsf{3.\;cos(A-B)+cos(A+B)=2\;cosA\;cosB}

\textsf{Here, A,B,C are angles of a triangle}

\mathsf{A+B+C=180^\circ}

\mathsf{Consider,}

\mathsf{cosA+cosB-cosC}

\textsf{Using formula (1) and (2), we get}

\mathsf{=2\,cos\left(\dfrac{A+B}{2}\right)\;cos\left(\dfrac{A-B}{2}\right)-\left(1-2\,sin^2\dfrac{C}{2}\right)}

\mathsf{=2\,cos\left(\dfrac{180^\circ-C}{2}\right)\;cos\left(\dfrac{A-B}{2}\right)-1+2\,sin^2\dfrac{C}{2}}

\mathsf{=2\,cos\left(90^\circ-\dfrac{C}{2}\right)\;cos\left(\dfrac{A-B}{2}\right)+2\,sin^2\dfrac{C}{2}-1}

\mathsf{=2\,sin\dfrac{C}{2}\;cos\left(\dfrac{A-B}{2}\right)+2\,sin^2\dfrac{C}{2}-1}

\mathsf{=2\,sin\dfrac{C}{2}\left[cos\left(\dfrac{A-B}{2}\right)+sin\dfrac{C}{2}\right]-1}

\mathsf{=2\,sin\dfrac{C}{2}\left[cos\left(\dfrac{A-B}{2}\right)+sin\left(\dfrac{180^\circ-(A+B)}{2}\right)\right]-1}

\mathsf{=2\,sin\dfrac{C}{2}\left[cos\left(\dfrac{A-B}{2}\right)+sin\left(90^\circ-\dfrac{A+B}{2}\right)\right]-1}

\mathsf{=2\,sin\dfrac{C}{2}\left[cos\left(\dfrac{A-B}{2}\right)+cos\left(\dfrac{A+B}{2}\right)\right]-1}

\mathsf{=2\,sin\dfrac{C}{2}\left[cos\left(\dfrac{A+B}{2}\right)+cos\left(\dfrac{A-B}{2}\right)\right]-1}

\textsf{Using formula (3), we get}

\mathsf{=2\,sin\dfrac{C}{2}\left[2\,cos\dfrac{A}{2}\,cos\dfrac{B}{2}\right]-1}

\mathsf{=4\,cos\dfrac{A}{2}\,cos\dfrac{B}{2}\,sin\dfrac{C}{2}-1}

\implies\boxed{\mathsf{cosA+cosB-cosC=-1+4\,cos\dfrac{A}{2}\,cos\dfrac{B}{2}\,sin\dfrac{C}{2}}}

\underline{\textbf{Find more:}}

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