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Which of the following is equal to tanθ+secθ-1tanθ- secθ+ 1\frac{\tanθ + \secθ - 1}{\tanθ - \secθ + 1} ?
निम्नलिखित में से कौनसा विकल्प tanθ+secθ-1tanθ- secθ+ 1\frac{\tanθ + \secθ - 1}{\tanθ - \secθ + 1} के बराबर है?
1+sinθcosθ\frac{1 + \sinθ}{\cosθ}
1+cosθsinθ\frac{1 + \cosθ}{\sinθ}
1+tanθcotθ\frac{1 + \tanθ}{\cotθ}
1+cotθtanθ\frac{1 + \cotθ}{\tanθ}
Correct Answer:
tanθ+secθ-1tanθ- secθ+ 1\frac{\tanθ + \secθ - 1}{\tanθ - \secθ + 1}
sinθcosθ+1cosθ-1sinθcosθ- 1cosθ+ 1\frac{\frac{\sinθ}{\cosθ} + \frac{1}{\cosθ} - 1}{\frac{\sinθ}{\cosθ} - \frac{1}{\cosθ} + 1}
sinθ- cosθ+ 1sinθ+cosθ- 1\frac{\sinθ - \cosθ + 1}{\sinθ + \cosθ - 1}
sinθ- cosθ+ 1sinθ+cosθ- 1\frac{\sinθ - \cosθ + 1}{\sinθ + \cosθ - 1} ×\times sinθ+ cosθ+ 1sinθ+cosθ+ 1\frac{\sinθ + \cosθ + 1}{\sinθ + \cosθ + 1}
1+2sinθ+sin2θ-cos2θ sin2θ+ cos2θ+2sinθcosθ-1\frac{1 + 2\sinθ + {\sin}^2θ - {\cos}^2θ}{ {\sin}^2θ + {\cos}^2θ+ 2\sinθ \cosθ - 1}
2sinθ+2sin2θ2sinθcosθ\frac{2\sinθ + 2{\sin}^2θ}{2\sinθ \cosθ}
2sinθ(1+sinθ) 2sinθcosθ\frac{2\sinθ (1 + \sinθ) }{2\sinθ \cosθ}
1+2sinθ+sinθ2-cosθ2 sinθ2+ cosθ2+2sinθcosθ-1\frac{1 + 2\sinθ + {\sinθ}^2 - {\cosθ}^2}{ {\sinθ}^2 + {\cosθ}^2 + 2\sinθ \cosθ - 1}
2sinθ+2sinθ2 2sinθcosθ\frac{2\sinθ + 2{\sinθ}^2 } { 2\sinθ \cosθ }
2sinθ(1+sinθ) 2sinθcosθ\frac{2\sinθ (1 + \sinθ) }{ 2\sinθ \cosθ }
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