Math, asked by scpanchal77, 19 days ago

sec⁴θ(1−sin⁴θ)−2tan²θ=1​

Answers

Answered by sudeshnna55
1

Answer:

L.H.S. = sec ⁴θ (1 - sin⁴θ) - 2 tan²θ

= sec⁴θ - sec⁴θsin⁴θ - 2tan²θ  

= sec⁴θ - sinθ⁴/cos⁴θ - 2tan²θ  

= sec⁴θ - tan⁴θ - 2tan²θ

= (sec²θ + tan²θ) (sec²θ - tan²θ) - - 2tan²θ[a² - b² = (a + b)(a - b)] [ Using identity  of tan²θ + 1 = sec²θ ]  

= (sec²θ + tan²θ) (1) - 2tan²θ  

= sec²θ - tan²θ  

=) 1 = RHS  [Using identity of tan²θ + 1 = sec²θ]

Step-by-step explanation:

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