Question

Prove that tan⁴O+ tan²O=sec⁴O-sec²O​

Answer 1

Hi, hope this helps:-)

Answer 2

To Prove :-

• tan⁴ θ + tan² θ = sec⁴ θ - sec²θ

Proof :-

So here we would be proving the L.H.S. equal to R.H.S.,

⇒ tan⁴ θ + tan² θ

Taking tan² θ as common,

⇒ tan² θ (tan² θ + 1)

Identity as we know that,

• sec² θ = 1 + tan²θ

By using it we gets,

⇒ tan² θ (sec² θ)

Now putting sec²θ - 1 in the place of tan²θ,

⇒ (sec²θ - 1) (sec² θ)

⇒ (sec²θ - 1) × (sec² θ)

⇒ sec⁴θ - sec² θ

Hence proved..!!!

Additional Information :

• sin² θ + cos² θ = 1
• sin² θ = 1 - cos²θ
• sec²θ = 1 + tan²θ
• cot²θ = cosec²θ - 1

Reciprocal identities :-

• sin θ = 1 / cosecθ
• cosec θ = 1 / sin θ
• cos θ = 1 / sec θ
• sec θ = 1 / cos θ
• tan θ = 1 / cot θ
• cot θ = 1 / tan θ