Question

cos (π-A)×cos(π÷2+A)×cos(-A)÷tan(π+A)×tan(3π÷2+A)×sin(2π-A)=cosA​

Answer 1

$\huge{ \boxed{ \mathfrak{question}}}$

$\frac{ \cos(\pi - \:a) \times \cos( \frac{\pi}{2} \: - \: a ) \: \times \cos( - a) }{ \tan(\pi + \: a) \times \: \tan( \frac{3\pi}{2} \: + \: a ) \: \times \: \sin(2\pi \: - a) }$

Some important identities :-

✧ In 1st quadrant all trigonometric functions are positive.

✧ In 2nd quadrant Sine and cosec are positive.

✧ In third quadrant tan and cot are positive.

✧ In Fourth quadrant cos and sec are positive .

→ When we crosses 180° or 360° The trigonometric function will remain same .

→ When we crosses 90° and 270° the function will change .

______________________

☆ Cos ( π - A) = - Cos A

☆ Cos ( 90 - A) = Sin A

☆ Cos ( - A) = Cos A

☆ tan ( π + A) = tan A

☆ tan ( 3π/2 + A) = - Cot A

☆ Sin ( 2π - A ) = - Sin A

Putting these values in the Question .

$\implies \: \frac{( - \cos(a)) \: \times \: \sin(a) \: \times \: \cos(a) }{ \tan(a) \: \times \:( - \cot(a) ) \: \times \:( - \sin(a) ) }$

☆ Tan A × Cot A = 1

$\implies \: \frac{ - \cos(a). \cos(a) . \sin(a) }{ - 1. - ( \sin a)}$

$\implies \: - { \cos(a) }^{2}$

Answer 2

Answer:

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