1b:I[1254,["7330","static/chunks/d3ac728e-78c3d5d7a486ed3a.js","5159","static/chunks/5159-48592f0bf1a47f9d.js","8175","static/chunks/8175-f84d59b9bd013e10.js","8167","static/chunks/8167-b25693a0140130f2.js","664","static/chunks/664-bceb0f356f9312eb.js","3703","static/chunks/app/practice-question/%5Bslug%5D/page-d1408559b663b83f.js"],"PracticeQuestionCard"] 1c:T52e,

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & \sqrt {{{\sec }^2}\theta + {\text{cose}}{{\text{c}}^2}\theta } \times \sqrt {{{\tan }^2}\theta - {{\sin }^2}\theta } \cr & = \sqrt {1 + {{\tan }^2}\theta + 1 + {{\cot }^2}\theta } \times \sqrt {{{\sin }^2}\theta \left( {\frac{1}{{{{\cos }^2}\theta }} - 1} \right)} \cr & = \sqrt {2 + {{\tan }^2}\theta + {{\cot }^2}\theta } \times \sqrt {\frac{{{{\sin }^2}\theta .{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \cr & = \left( {\tan \theta + \cot \theta } \right) \times \left( {\frac{{{{\sin }^2}\theta }}{{\cos \theta }}} \right) \cr & = \left( {\frac{{\sin \theta }}{{\cos \theta }} + \frac{{\cos \theta }}{{\sin \theta }}} \right) \times \left( {\frac{{{{\sin }^2}\theta }}{{\cos \theta }}} \right) \cr & = \frac{1}{{\sin \theta .\cos \theta }} \times \frac{{{{\sin }^2}\theta }}{{\cos \theta }} \cr & = \sin \theta .{\sec ^2}\theta \cr} $$

1d:T425,Answer & Solution Answer: Option B Solution: $$\eqalign{ & \sqrt {{{\sec }^2}\theta + {\text{cose}}{{\text{c}}^2}\theta } \times \sqrt {{{\tan }^2}\theta - {{\sin }^2}\theta } \cr & = \sqrt {1 + {{\tan }^2}\theta + 1

The value of ^2 + cose c^2 × ^2 - ^2 is equal to:

Solution