Question

Prove that, tan^4∅ + tan^2∅ = sec^4∅ - sec^2∅

Answer 1

Answer:

CORRECT QUESTION : tan(Ø)⁴ + tan(Ø)² = sec(Ø)⁴ - sec(Ø)²

SOLUTION :

$tan(\theta)^{4} + tan(\theta)^{2} = sec(\theta)^{4} - sec(\theta)^{2}$

$\dfrac{sin \theta}{cos \theta}^{4} + \dfrac {sin \theta}{cos \theta}^{2}$ = $\dfrac {1}{cos \theta}^{4} - \dfrac {1}{cos \theta}^{2}$

$\dfrac {sin(\theta)^{2} + cos(\theta)^{2} sin(\theta)^{2} -1 + cos(\theta)^{2}}{cos(\theta)^{4}}$

$\dfrac {[sin (\theta)^{2} + cos (\theta)^{2}] \times sin(\theta)^{2} - 1-cos(\theta)^{2}}{cos(\theta)^{4}}$

$\dfrac {1 sin(\theta)^{2} - sin(\theta)^{2}}{cos(\theta)^{4}}$

$\dfrac {0}{cos(\theta)^{4}}$

→ $0 \: = \: 0$

→ $\boxed{0 \: = \: 0}$