The value of the expression 1 - frac{sin^2 y} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given expression: $E = 1 - \frac{\sin^2 y}{1 + \cos y} + \frac{1 + \cos y}{\sin y} - \frac{\sin y}{1 - \cos y}$Step $1$: Simplify the first two te

Vedclass · Accepted Answer

$\cos y$

Vedclass · Answer

(D) Given expression: $E = 1 - \frac{\sin^2 y}{1 + \cos y} + \frac{1 + \cos y}{\sin y} - \frac{\sin y}{1 - \cos y}$Step $1$: Simplify the first two terms:$1 - \frac{\sin^2 y}{1 + \cos y} = \frac{1 + \cos y - (1 - \cos^2 y)}{1 + \cos y} = \frac{\cos y + \cos^2 y}{1 + \cos y} = \frac{\cos y(1 + \cos y)}{1 + \cos y} = \cos y$Step $2$: Simplify the last two terms:$\frac{1 + \cos y}{\sin y} - \frac{\sin y}{1 - \cos y} = \frac{(1 + \cos y)(1 - \cos y) - \sin^2 y}{\sin y(1 - \cos y)} = \frac{1 - \cos^2 y - \sin^2 y}{\sin y(1 - \cos y)} = \frac{\sin^2 y - \sin^2 y}{\sin y(1 - \cos y)} = 0$Step $3$: Combine the results:$E = \cos y + 0 = \cos y$

The value of the expression $1 - \frac{\sin^2 y}{1 + \cos y} + \frac{1 + \cos y}{\sin y} - \frac{\sin y}{1 - \cos y}$ is equal to

Similar Questions

If $\sin x + \sin^2 x = 1,$ then $\cos^8 x + 2\cos^6 x + \cos^4 x = $

If $A, B, C, D$ are the angles of a cyclic quadrilateral,then $\cos A + \cos B + \cos C + \cos D = $

Evaluate the sum: $\sin^2(3^{\circ}) + \sin^2(6^{\circ}) + \sin^2(9^{\circ}) + \dots + \sin^2(84^{\circ}) + \sin^2(87^{\circ}) + \sin^2(90^{\circ})$

If $\sin (\alpha+\beta)=5 \sin (\alpha-\beta)$,then $\frac{\sin 2 \beta}{5-\cos 2 \beta}=$

If $f_n(x) = \frac{1}{2n} [\sin^{2n} x + \cos^{2n} x]$,then $f_1(x) + f_2(x) - f_3(x) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module