Question

Prove the identify

(sec A - cos A) · (cot A + tan A) = tanA · sec A​

Answer 1

Step-by-step explanation:

Given:

• (sec A - cos A) · (cot A + tan A) = tanA · sec A

To Prove:

• LHS = RHS

Proof: Taking LHS first.

➮ (sec A - cos A) × (cot A + tan A)

➮ (1/cos A – cos A) (cos A/sin A + sin A/cos A)

➮ ( 1 – cos² A/cos A) (cos² A + sin² A/sinA cos A)

➮ sin² A/cos A × 1/sin A cos A

➮ sin A/cos A × 1/cos A

➮ tan A × sec A

Hence, LHS = RHS proved.

★ Identities used here ★

• sec θ = 1/cosθ

• cot θ = cosθ/sinθ

• tan θ = sinθ/cosθ

• sin² θ = 1 – cos²θ

• cos²θ + sin²θ = 1

Answer 2

To Prove:

$\bull\sf{(sec A - cos A) · (cot A + tan A) = tanA · sec A}$

Proof:

LHS

$\leadsto \sf \: (secA - cosA).(cotA + tanA) \: \green\bigstar\\ \\ \leadsto \sf \: ( \frac{1}{cosA} - cosA).( \frac{cosA}{sinA} + \frac{sinA}{cosA} ) \\ \\ \leadsto \sf \: ( \frac{1 - cos {}^{2}A }{cosA} ).( \frac{cos {}^{2}A + sin { }^{2} A }{sinA.cosA} ) \\ \\ \leadsto \sf \: ( \frac{sin {}^{ \cancel{2}}A }{cosA} ).( \frac{1}{ \cancel{sinA}.cosA} ) \\ \\ \leadsto \sf \: ( \frac{sinA}{cosA} ).( \frac{1}{cosA} ) \\ \\ \leadsto \sf \: tanA.secA \: \red \bigstar$

RHS