Question

if sec ^2 ø (1+sin ø) (1-sin ø) = K. find the value of K​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Value\:of\:k=1}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline{\bold{Given :}}} \\ \tt: \implies {sec}^{2} \: \theta(1 + sin \: \theta)(1 - sin \: \theta) = k \\ \\ \red{\underline{\bold{To \: Find :}}} \\ \tt: \implies k =?$

• According to given question :

$\bold{As \: we \: know \: that} \\\tt: \implies k = {sec}^{2} \: \theta(1 + sin \: \theta)(1 - sin \: \theta) \\ \\ \tt \green\star \: sec \: \theta = \frac{1}{cos \: \theta} \\ \\ \tt: \implies k = \frac{1}{ {cos}^{2} \: \theta } (1 + sin \: \theta)(1 - sin \: \theta) \\ \\ \tt \green \star \: {cos}^{2} \: \theta = 1 - {sin}^{2} \: \theta \\ \\ \tt: \implies k = \frac{1}{1 - {sin}^{2} \: \theta} (1 + sin \: \theta)(1 - sin \: \theta) \\ \\ \tt \green \star\: {a}^{2} - {b}^{2} = (a + b)(a - b) \\ \\ \tt: \implies k = \frac{1}{ {1}^{2} - {sin}^{2} \: \theta}(1 + sin \: \theta)(1 - sin \: \theta) \\ \\ \tt: \implies k = \frac{(1 + sin \: \theta)(1 - sin \: \theta)}{ (1 + sin \: \theta)(1 - sin \: \theta)} \\ \\ \green{\tt: \implies k = 1} \\ \\ \green{\tt \therefore Value \: of \: k \: is \: 1}$

Answer 2

Hello !

k = sec²o(1+sino)(1-sino)

K = 1/cos²o (1+sino)(1-sino) ...,............... as sec²o = 1/cos²o

K = 1/1-sin²o(1+sino)(1-sino) ................... as cos²o = 1-sin²o

K = 1/(1+sino)(1-sino)[(1+sino)(1-sino) ....... 1-sin²o = (1+sino)(1-sino)

hence, k = 1.