Question

sec-tan/sec+tan =1+2tan²-2sec tan
​

Answer 1

Step-by-step explanation:

by simplification you can get it

Answer 2

Proved!

Step-by-step explanation:

We are provided that;

⇒ secA - tanA/secA + tanA = 1 + 2tan²A - 2secA × tanA

Before moving onto the problem, let's recall the trigonometric and algebraic identities. These three identities are important for us to solve this problem:

→ sec²A - tan²A = 1 (or) sec²A = 1 + tan²A

→ (a + b)(a - b) = a² - b²

→ (a - b)² = a² + b² - 2ab

Now, from the given information, let's take LHS and solve the problem.

⇒ secA - tanA/secA + tanA

⇒ (secA - tanA)(secA - tanA)/(secA + tanA)(secA - tanA)

⇒ (secA - tanA)²/(sec²A - tan²A)

⇒ [sec²A + tan²A - (2 × secA × tanA)]/1

⇒ [sec²A + tan²A - 2 × secA × tanA]

⇒ [(tan²A + 1) + tan²A - 2 × secA × tanA]

⇒ [1 + 2tan²A - 2 × secA × tanA]

As we can observe that, LHS = RHS, Hence, proved!