Recent content by surajkrishna

Thanks for the help.

$\displaystyle \frac{1+ \sin \theta}{\sqrt {1 - \sin^{2} \theta}}= x $ After this how do you get this: $\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}=x \implies \sin \theta = \frac{x^{2}-1}{x^{2}+1}$

Q2) If sec\theta + tan\theta = x, then sin\theta is equal to a) x2+1/2x b) x2+1/x2-1 c) x2-1/x2+1 d) 2x/x2​-1

I have some questions and doubts in trigonometry. I hope somebody can solve these questions. Q1) If for real values of x, cos\theta = x +\frac{1}{x}, then a) \theta is acute angle b) \theta is right angle c) \theta is an obtuse angle d) no value of \theta is possible I will post the following...