Doubt by Ziya
Solution :
(k+1)=sec²θ(1-sinθ)(1+sinθ)
(k+1)=[1/cos²θ]×[1-sinθ)(1+sinθ] [∵secθ=1/cosθ]
(k+1)=[1/cos²θ]×[1²-sin²θ] [∵(a-b)(a+b)=(a²-b²)]
(k+1)=[1/cos²θ]×[cos²θ]
(k+1)=[1/cos²θ]×[cos²θ]
(k+1)=1
k+1=1
k=1-1
k=0
Hence, the value of k is 0