Question

prove that
sec⁴ theta- tan⁴ theta= 1+2 tan² theta​

Answer 1

Required Answer:-

Given to prove:

• sec⁴(x) - tan⁴(x) = 1 + 2tan²(x)

Proof:

Taking LHS,

sec⁴(x) - tan⁴(x)

= [sec²(x)]² - [tan²(x)]²

= (sec²(x) + tan²(x))(sec²(x) - tan²(x))

We know that,

➡ sec²(x) - tan²(x) = 1

So,

(sec²(x) + tan²(x))(sec²(x) - tan²(x))

= (sec²(x) + tan²(x)) × 1

= sec²(x) + tan²(x)

As sec²(x) - tan²(x) = 1,

So, sec²(x) = 1 + tan²(x)

So,

sec²(x) + tan²(x)

= 1 + tan²(x) + tan²(x)

= 1 + 2 tan²(x)

Taking RHS,

= 1 + 2tan²(x)

Hence, LHS = RHS (Proved)

Therefore, sec⁴(x) - tan⁴(x) = 1 + 2 tan²(x)

Formula Used:

• sec²(x) - tan²(x) = 1

Other Formula:

• sin²(x) + cos²(x) = 1
• cosec²(x) - cot²(x) = 1
• sin(90° - x) = cos(x)
• cosec(90° - x) = sec(x)
• tan(90° - x) = cot(x)

Answer 2

$\sf{\sec^4\left(\theta\right)-\tan ^4\left(\theta\right)}$

$\sf{=\left(\sec ^2\left(\theta\right)\right)^2-\left(\tan ^2\left(\theta\right)\right)^2}$

$\sf{=\left(\sec^2\left(\theta\right)+\tan ^2\left(\theta\right)\right)\left(\sec ^2\left(\theta\right)-\tan ^2\left(\theta\right)\right)}$

$\sf{=\left(\sec ^2\left(\theta\right)+\tan ^2\left(\theta\right)\right)\left(\sec \left(\theta\right)+\tan \left(\theta\right)\right)\left(\sec \left(\theta\right)-\tan \left(\theta\right)\right)}$

$\sf{=\sec ^2\left(\theta\right)+\tan ^2\left(\theta\right)}$

$\sf{=1+\tan^2\left(\theta\right)+\tan ^2\left(\theta\right)}$

$\sf{=1+2\tan^2\left(\theta\right)}$

Hence, LHS = RHS (Proved)