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\operatorname { sin } ( 40 ^ { \circ } + \theta ) \operatorname { cos } ( 10 ^ { \circ } + \theta ) - \operatorname { cos } ( 40 ^ { \circ } + \theta ) \operatorname { sin } ( 10 ^ { \circ } + \theta ) = \frac { 1 } { 2 } |
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Answer» Here,= Sin (40 + A) Cos (10 + A) - cos (40 + A) Sin (10 + A)As, sin A cos B - cos A sin B = sin (A - B)so,= sin (40 + A - 10 - A)= sin 30= 1/2 = RHSHence, proved. here, for convenience we used theta as A |
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