Question

if k+1 =sec²Φ (1+sinΦ) (1-sinΦ) then find the value of k​

Answer 1

k + 1 = sec^2€ (1+sin€)(1-sin€)

k + 1 = sec^2€ ( 1 - sin^2€ )

k + 1 = sec^2€ (cos^2€)

k +1 = sec^2€ ×1/sec^2€

k + 1 = 1

k = 1 - 1

k = 0........●

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Hope it was helpful. ^_^

Answer 2

$k + 1 = { \sec {}^{2} ( \alpha ) } (1 - { \sin {}^{2} ( \alpha ) }) \\ k + 1 = { \sec( \alpha ) }^{2}. { \cos {}^{2} ( \alpha ) } \\ k + 1 = 1 \\ k = 0$