Question

If k+1=sec^2 (1+sin)(1-sin),then find the value of k

Answer 1

Given

K + 1 = sec²∅(1 + sin∅)(1 - sin∅)

→ K + 1 = sec²∅(1 - sin²∅)

→ K + 1 = sec²∅.cos²∅

→ K + 1 = 1

→ K = 0

Answer 2

$\Large{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

k = 0

$\large{\underline{\underline{\red{\mathfrak{Step-By-Step-Explanation :}}}}}$

k + 1 = Sec²A (1 + SinA)(1 - SinA)

⇒k + 1 = Sec²A [(1)² - (SinA)²]

${\boxed{\tt{As, \: (a \: + \: b) (a \: - \: b) \: = \: a^2 \: - \: b^2}}}$

⇒k + 1 = Sec²A [1 - Sin²A]

As we know,

$\Large{\boxed{\sf{1 \: - \: Sin^2A \: = \: Cos^2A}}}$

Put Values

⇒k + 1 = Sec²A(Cos²A)

As we know,

$\Large{\boxed{\sf{SecA \: = \: \frac{1}{Cos}}}}$

⇒k + 1 = Sec²A(1/Sec²A)

⇒k + 1 = Sec²A/Sec²A

⇒k + 1 = 1

⇒k = 1 - 1

⇒k = 0

$\LARGE \implies {\boxed{\boxed{\sf{k \: = \: 0}}}}$

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