Question

Prove that : sin A (1 + tan A) + cos A (1 + cot A) = sec A + cosec A​

Answer 1

Step-by-step explanation:

Solution :-

On taking LHS

=> sin A (1 + tan A) + cos A (1 + cot A)

=>sinA[1+(sinA/cosA)]+cosA[1+(cosA/sinA)]

Since , Tan A = Sin A / Cos A

and Cot A = Cos A / Sin A

=>sinA[(cosA+sinA)/cosA]+cosA[(sinA+cosA)/sinA]

=>[sinA(cosA+sinA)/cosA]+[cos A(sinA+cosA)/sinA]

=> (cosA+sinA)[(sin A/cos A)+(cos A/sin A)]

=> (cosA+sinA)[(sin²A+cos²A)/(sinA cosA)]

=> (cosA+sinA)[1/(sin A cos A)]

Since , Sin² A + Cos² A = 1

=> (cos A + sin A )/(sin A cos A)

=> [cosA/(sinAcosA)]+[sinA/(sinA cosA)]

=> (1/sin A) + (1/cos A)

=> cosec A + sec A

=> sec A + cosec A

=> RHS

=> LHS = RHS

Hence, Proved.

Answer:-

[sinA(1+tanA)]+[cosA(1+cotA)] =secA+cosecA

Used formulae :-

→ Tan A = Sin A / Cos A

→ Cot A = Cos A / Sin A

→ Sin² A + Cos² A = 1

→ Sec A = 1/Cos A

→ Cosec A = 1/Sin A