Question

heyaaaa please solve this question


don't post irrelevant answer

need errorless answer​

Answer 1

in ∆ ABC ,

A+B+C = 180° = π rads

=> A+B = π-C => cos (A+B) = cos (π-C) = - cos C

=> sin (A+B) = sin (π-C) = sin C

=> B+C = π-A => cos (B+C) = cos (π-A) = -cosA

=> sin (B+C) = sin (π-A) = sin A

PT => 1) cos A cos C + (-cos C)(-cos A) = 2 cos A cos C

and cos A sin C - (sin C)(-cos A) = 2 cos A sin C

=> divide, we get ,

=> LHS = 2 cos A cos C / 2 cos A sin C

=> cot C = RHS

Formula used => sin ( π - A ) = sin A and

cos (π-A) = - cos A

2)))

similarly , tan (A+B) = tan (π-C) = -tan C

tan (B+C) = tan (π-A) = -tanA

tan (C+A) = tan (π-B) = -tan B

and tan (π-A) = - Tan A ,

tan (2π - B) is an angle in the fourth quadrant where tan of an angle is negative .

=> tan (2π - B)= -tan B

and tan (3π - C) is an angle in second quadrant where the tangent of an angle is negative

=> tan (3π-C) = -tan C

now ,

=> LHS => (- tan C - tan B - tan A )/( - tan A - tan B - tan C )

=> 1 = RHS

Formulas used => tan ( π- A) = - tan A