Question

More than One Answer Type
एक से अधिक उत्तर प्रकार के प्रश्न

If $x=\left(\frac{{\sin }^2\theta {\sec }^2\theta +1}{{\sec }^2\theta }\right)+\left(\frac{{\cot }^2\theta }{1+cosec\theta }\right),$$y=\frac{1+{\cos }^2\theta cose{c}^2\theta }{cose{c}^2\theta }+\frac{{\tan }^2\theta }{1+\sec \theta }$ and $z=\frac{1-{\sin }^2\theta +\sin \theta \cos \theta }{\sin \theta -{\sin }^3\theta },$ where 0° < θ < 90°, then the correct option(s) is/are

यदि $x=\left(\frac{{\sin }^2\theta {\sec }^2\theta +1}{{\sec }^2\theta }\right)+\left(\frac{{\cot }^2\theta }{1+cosec\theta }\right),$$y=\frac{1+{\cos }^2\theta cose{c}^2\theta }{cose{c}^2\theta }+\frac{{\tan }^2\theta }{1+\sec \theta }$ तथा $z=\frac{1-{\sin }^2\theta +\sin \theta \cos \theta }{\sin \theta -{\sin }^3\theta },$ जहाँ 0° < θ < 90°, तब सही विकल्प है/हैं

• A. x² + y² = x²y²
• B. z = xy(sinθ + cosθ)
• C. x + y + z = 0
• D. x + y = z

Answer 1

Answer: (A), (B), (D)

x + y = z

Solution