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(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
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