If (1+sinθ)/cosθ = x, then find the value of

Solution

(1+sinθ)/cosθ = x sinθ/cosθ  + 1/cosθ = x tanθ + secθ =  x  ...........................(i) secθ =  x – tanθ On squaring both sides, (secθ)² = (x-tanθ)² sec²θ = x²+ tan²θ-2xtanθ sec²θ-tan²θ =  x² -2xtanθ 1 = x² -2xtanθ tanθ = (x²-1)/2x Now putting the value of tanθ in equation (i) (x²-1)/2x + secθ =  x   secθ =  x  - (x²-1)/2x secθ = (2x²-x²+1)/2x =  (x²+ 1)/2x Alternate method: (1+sinθ)/cosθ = x sinθ/cosθ  + 1/cosθ = x tanθ + secθ =  x  secθ+tanθ =  x     ...........................(i) So  secθ-tanθ =  1/x     ...........................(ii)    {As sec²θ - tan² θ =  1 } Adding both equations 2secθ = x + 1/x  = (x²+1)/x So secθ = (x²+1)/2x