(1-sinxtanx)/(1+secx)=secx

Question

Please Help Me To Solve That Question

Answer 1

First, we need to find out whether this is an identity (to prove) or an equation (to solve). Let x=0:

(1-sin(0)tan(0))/(1+sec(0))=½≠sec(0) because sec(0)=1. So it's an equation that has to be solved for x. (1-sin(x)tan(x))/(1+sec(x))-sec(x)=0 has to be solved.

(1-sin²(x)/cos(x))/(1+1/cos(x))=1/cos(x),

(cos(x)-sin²(x))/(cos(x)+1)=1/cos(x),

(cos²(x)+cos(x)-1)/(cos(x)+1)=1/cos(x).

Let y=cos(x):

(y²+y-1)/(y+1)=1/y,

y(y²+y-1)=y+1 (cross-multiplication),

y³+y²-y=y+1,

y³+y²-2y-1=0.

This cubic has three solutions, but only one is within range for cosine. 

Let f(y)=y³+y²-2y-1, then f'(y)=3y²+2y-2, and use Newton's Method to find the solution:

y_n+1=y_n-f(y_n)/f'(y_n). Start with y=0, then y₁=-½, and eventually arrive at y=-0.4450 approx. cos(x)=y so x=116.426° approx.

This is one solution to the question for x between 0 and 360°. The other is 360-116.426=243.574°. However, I think the question has been misstated, because the solution (which does satisfy the original equation) is too complicated.