Question

a trigonometric problem pls help​

Answer 1

Question: (sec^6 A + tan^2 A)/(sec^2A + tan^2A) = 1 + sec^2 A tan^2 A

Step-by-step explanation:

$\implies \sf{\dfrac{sec^6\theta+tan^2 \theta}{sec^2 \theta+tan^2 \theta} }\\\\\\\implies \sf{\dfrac{(sec^2\theta)^3 + (tan^2\theta)^3}{sec^2\theta+ tan^2\theta} }\\\\\\\implies\sf{\dfrac{(sec^2\theta+tan^2\theta)((sec^2\theta)^2 + (tan^2\theta)^2 - \sec^2\theta \tan^2\theta)}{(sec^2\theta + tan^2\theta) } }\\\\\\\implies\sf{(sec^2\theta)^2 +(tan^2\theta)^2 - 2sec^2\theta tan^2\theta + sec^2 \theta tan^2\theta}\\\\\implies \sf{(sec^2\theta - tan^2\theta)^2 + sec^2\theta \tan^2\theta}$

$\implies \sf{ 1 + sec^2\theta tan^2 \theta}$

  Solved using:

• a³ + b³ = (a + b)(a² + b² - ab)
• a² + b² - 2ab = (a - b)²
• sec²A - tan²A = 1

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Answer 2

for further explanation please refer the above attachment