যদি \(sinθ – cosθ = 0 (0°≤ θ ≤ 90°)\) এবং \(secθ + cosecθ = x\) হয়, তাহলে \(x\) -এর মান (a) 1 (b) 2 (c) \(\sqrt2\) (d) \(2\sqrt2\)
Answer: D
\(sinθ-cosθ=0 \)
বা, \((sinθ-cosθ)^2=0 \)
বা, \(sin^2θ+cos^2θ-2sinθcosθ=0 \)
বা, \(1-2sinθcosθ=0 \)
বা, \(2sinθcosθ=1 \)
বা, \(sinθcosθ=\cfrac{1}{2} \)
আবার, \((sinθ+cosθ)^2\)
\(=sin^2θ+cos^2θ+2 sinθcosθ \)
বা, \((sinθ+cosθ)^2=1+2×\cfrac{1}{2} \)
বা, \((sinθ+cosθ)^2=2 \)
বা, \(sinθ+cosθ=\sqrt2\)
\(∴secθ+cosecθ=x \)
বা, \(x=secθ+cosecθ\)
\(=\cfrac{1}{cosθ}+\cfrac{1}{sinθ}\)
\(=\cfrac{sinθ+cosθ}{sinθcosθ}=\cfrac{√2}{\cfrac{1}{2}}=2\sqrt2\)
\(sinθ-cosθ=0 \)
বা, \((sinθ-cosθ)^2=0 \)
বা, \(sin^2θ+cos^2θ-2sinθcosθ=0 \)
বা, \(1-2sinθcosθ=0 \)
বা, \(2sinθcosθ=1 \)
বা, \(sinθcosθ=\cfrac{1}{2} \)
আবার, \((sinθ+cosθ)^2\)
\(=sin^2θ+cos^2θ+2 sinθcosθ \)
বা, \((sinθ+cosθ)^2=1+2×\cfrac{1}{2} \)
বা, \((sinθ+cosθ)^2=2 \)
বা, \(sinθ+cosθ=\sqrt2\)
\(∴secθ+cosecθ=x \)
বা, \(x=secθ+cosecθ\)
\(=\cfrac{1}{cosθ}+\cfrac{1}{sinθ}\)
\(=\cfrac{sinθ+cosθ}{sinθcosθ}=\cfrac{√2}{\cfrac{1}{2}}=2\sqrt2\)
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