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