Question

(ii) (1 + cot A + tan A) (sin A -cos A) = sin A tan A- cos A cot A.
​

Answer 1

1. ( 1 + Cot A + Tan A ) (sin A - cos A) = sin A tan A - cos A cot A

Taking LHS

→ (1 + cot A + Tan A ) (Sin A - cos A)

Opening the brackets :

→ 1(Sin A - cos A) + cot A ( SinA + cos A ) + Tan A ( Sin A - cos A)

→ sin A - cos A + cot A sin A + cot A cos A + Tan A sin A - tan A cos A

Now we know that

• cot A = cos A / sin A
• tan A = sin A / cos A

$\tt sin A - cos + \dfrac{cos\ A}{sin\ A}\ sin A + \\ \\ \tt cot A cos A + tan A sin A - \dfrac{ sin\ A}{cos\ A}\ cos A$

→ sin A - cos A + cos A - cot A cos A + tan A sin A - sin A

→ sin A - sin A + cos A - cos A + tanA sin A - cot A cos A

→ Tan A sin A - cot A cos A

→ sin A tan A - cos A cot A

$\therefore$ LHS = RHS

Hence proved