$ \theta = \sin^{-1} \left( \frac{3}{5} \right) \text{ হলে, } \frac{1-\tan^{2} \theta}{1+\tan^{2} \theta} = ? \left( \text{if } \theta = \sin^{-1} \left( \frac{3}{5} \right) \text{ then } \frac{1-\tan^{2} \theta}{1+\tan^{2} \theta} = ? \right)$

দেওয়া আছে,
θ = sin^-1(3/5)
বা, sinθ = 3/5
আমরা জানি, cosθ = √(1 - sin²θ)
= √(1 - (3/5)²)
= √(1 - 9/25)
= √(16/25) = 4/5
∴ tanθ = sinθ / cosθ = (3/5) / (4/5) = 3/4

প্রদত্ত রাশি = (1 - tan²θ) / (1 + tan²θ)
= (1 - (3/4)²) / (1 + (3/4)²)
= (1 - 9/16) / (1 + 9/16)
= ((16-9)/16) / ((16+9)/16)
= (7/16) / (25/16)
= 7/25

শর্টকাট: আমরা জানি, cos2θ = (1 - tan²θ) / (1 + tan²θ)
আবার, cosθ = 4/5.
তাহলে cos2θ = 2cos²θ - 1 = 2(16/25) - 1 = 32/25 - 1 = 7/25.