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Prove That: Cos2 a + Cos2 B − 2 Cos a Cos B Cos (A + B) = Sin2 (A + B)

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प्रश्न

Prove that:
cos2 A + cos2 B − 2 cos A cos B cos (A + B) = sin2 (A + B)

संक्षेप में उत्तर
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उत्तर

\[\text{ LHS }= \cos^2 A + \cos^2 B - 2\cos A \cos B \cos\left( A + B \right)\]
\[ = \cos^2 A + 1 - \sin^2 B - 2\cos A \cos B \cos\left( A + B \right)\]
\[ = 1 + \cos^2 A - \sin^2 B - 2\cos A \cos B \cos\left( A + B \right)\]
\[ = 1 + \cos^2 A - \sin^2 B - 2\cos A \cos B \cos\left( A + B \right)\]
\[ = 1 + \cos\left( A + B \right)\cos\left( A - B \right) - 2\cos A \cos B \cos\left( A + B \right)\]
\[ = 1 + \cos\left( A + B \right)\left\{ \cos\left( A - B \right) - 2\cos A \cos B \right\}\]
\[ = 1 + \cos\left( A + B \right)\left( \cos A \cos B + \sin A \sin B - 2\cos A \cos B \right)\]
\[ = 1 + \cos\left( A + B \right)\left( - \cos A \cos B + \sin A \sin B \right)\]
\[ = 1 - \cos\left( A + B \right)\left( \cos A \cos B - \sin A \sin B \right)\]
\[ = 1 - \cos\left( A + B \right)\cos\left( A + B \right)\]
\[ = 1 - \cos^2 \left( A + B \right)\]
\[ = \sin^2 \left( A + B \right)\]
 = RHS
Hence proved.

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अध्याय 7: Values of Trigonometric function at sum or difference of angles - Exercise 7.1 [पृष्ठ २०]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.1 | Q 16.5 | पृष्ठ २०

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