যদি \(sin x = m sin y\) এবং \(tan x = n tan y\) হয় তবে দেখাও যে \(cos^2 x =\cfrac{m^2-1}{n^2-1}\)
Madhyamik 2025
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\(sin x = m sin y\)
বা, \(sin^2x=m^2sin^2y\)
বা, \(1-cos^2x=m^2(1-cos^2y)\)
বা, \(1-cos^2x=m^2-m^2cos^2y\)
বা, \(m^2cos^2y=m^2-1+cos^2x\)
বা, \(cos^2y=\cfrac{m^2-1+cos^2x}{m^2}-----(i)\)
\(tan x = n tan y\)
বা, \(\cfrac{sin x}{cos x}=n\cfrac{sin y}{cos y}\)
বা, \(\cfrac{m sin y}{cos x}=n\cfrac{sin y}{cos y}\) [\(sinx=m siny\) বসিয়ে পাই]
বা, \(\cfrac{m}{cos x}=\cfrac{n}{cos y}\)
বা, \(cos y=\cfrac{n. cos x}{m}\)
বা, \(cos^2 y=\cfrac{n^2 cos^2 x}{m^2}-----(ii)\)
\((i)\) এবং \((ii)\) নং সমীকরন থেকে পাই,
\(\cfrac{m^2-1+cos^2x}{m^2}=\cfrac{n^2 cos^2 x}{m^2}\)
বা, \(m^2-1+cos^2x=n^2 cos^2 x\)
বা, \(n^2 cos^2 x=m^2-1+cos^2x\)
বা, \(n^2 cos^2 x-cos^2x=m^2-1\)
বা, \(cos^2 x(n^2 -1)=m^2-1\)
বা, \(cos^2 x=\cfrac{m^2-1}{n^2 -1}\) [প্রমানিত]
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