Question

\(\frac{{sin\theta [\left( {1 - tan\theta } \right)tan\theta + {{\sec }^2}\theta ]}}{{\left( {1 - \sin \theta } \right)\tan \theta \left( {1 + tan\theta } \right)\left( {sec\theta + tan\theta } \right)}}\) is equal to:
1. sinθ cosθ 
2. -1
3. cosecθ secθ 
4. 1

Answer 1

Correct Answer - Option 4 : 1

GIVEN:

\(\frac{{sinθ [\left( {1 - tanθ } \right)tanθ + {{\sec }^2}θ ]}}{{\left( {1 - \sin θ } \right)\tan θ \left( {1 + tanθ } \right)\left( {secθ + tanθ } \right)}}\)

FORMULA USED:

\({\sec ^2}θ - \;{\tan ^2}θ = 1\), sin²θ + cos²θ = 1. tanθ = sinθ/cosθ, secθ = 1/cosθ 

CALCULATION:

\(\frac{{sinθ [\left( {1 - tanθ } \right)tanθ + {{\sec }^2}θ ]}}{{\left( {1 - \sin θ } \right)\tan θ \left( {1 + tanθ } \right)\left( {secθ + tanθ } \right)}}\)

\( ⇒ \;\frac{{\sin θ \;[\tan θ - \;{{\tan }^2}θ + \;1 + \;{{\tan }^2}θ ]}}{{(1 - \;\sin θ )\;\tan θ \;(1 + \;\tan θ )\;(\sec θ + \;\tan θ )}}\)

\( ⇒ \frac{{\sin θ \;(1 + \;\tan θ )}}{{(1 - \sin θ )\;\tan θ \;(1 + \;\tan θ )\;(\sec θ + \;\tan θ )}}\)

\(⇒ \;\frac{{\sin θ }}{{(1 - \;\sin θ )\;\frac{{\sin θ }}{{\cos θ }}\;(\sec θ + \;\tan θ )}}\)

\(⇒ \;\frac{{\cos θ }}{{(1 - \;\sin θ )\;\left( {\frac{1}{{\cos θ }} + \;\frac{{\sin θ }}{{\cos θ }}} \right)}}\)

\( ⇒ \frac{{{{\cos }^2}θ }}{{(1 - \sin θ )\;(1 + \;\sin θ )}}\)

\(⇒ \frac{{{{\cos }^2}θ }}{{(1 - \;{{\sin }^2}θ )}}\)

\(⇒ \;\frac{{{{\cos }^2}θ }}{{{{\cos }^2}θ }}\)

⇒ 1