Question

In a triangle ABC, sec A (cos B cos o - sin B sin C) is equal to_______​

Answer 1

Explanation:

In a triangle ABC ,

A + B + C = 180°

I.e . B+C = 180 - A ---------------- ( 1 ) .

now , sec A ( cos B cos C - sin B sin C ) . ( 2 )

As we know that

cos ( x + y ) = cos X cos Y - sin X sin Y

Similarly , cos B cos C - sin B sin C = cos ( B + C )

put in equation. ( 2 )

Sec A ( cos B cos C - sin B sin C ) .

Sec A [ cos ( B + C ) ] .

put the value of B + C from equation ( 2 ) .

Sec A cos [ ( 180 - A ) ] .

Sec A [ cos ( - A ) ] because cos ( 180 - A ) = cos ( - A ) .

Sec A ( cos A ) .

Ans = 1 .

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