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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
46.2 × 8.4 × 3 + ? = 1200
63- [22-{24 ÷ 3-(9-15 ÷ 5) ÷ 6}]=?
14 × 6 + 9 × 11 = (82 – 3) × ?
(12% of 1250 + 85% of 400) x 10 = ?2
∛21952 × 44 = ? × 14
`sqrt(5476)` + 40% of 1640 = ?`xx` 4 - 2020
654.056 + 28.9015 × 44.851 – 43.129 = ?
212 + 14 × 23 – 28 × 15 = ?
?3 - 25 × 11 = 30 - 15 × 12
