Question

please solve the oues

Answer 1

Hey !!!

Sec²¢ ( 1 + sin¢ ) ( 1 - sin¢ ) = k

=> sec²¢ {( 1 ² - sin²¢ ) } = k

=> sec²¢ × cos²¢ = k

[ °•° ( 1 - sin²¢ = cos²¢ ) ]

=> 1/cos²¢ × cos²¢ = k

[ °•° ( sec²¢ = 1/cos²¢) ]


1 = k

value of k = 1 Answer ✔

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Hope it helps you !!!

@Rajukumar111 ☺

Answer 2

$hello \: mate... \\ gievn \\ \sec {}^{2} (theeta) (1 + \sin(theeta) )(1 - sin(theeta)) = k \\ sec {}^{2}(theeta)(1 - sin {}^{2} (theeta)) = k \\ sec {}^{2} (theeta)(cos {}^{2} (theeta)) = k \\ sec {}^{2} (theeta)( \frac{1}{sec {}^{2} (theeta)} ) = k \\ 1 = k \\ k = 1. \\ \\ hope \: it \: helps \: u \: dear....$