It's trigo time #4

Geometry Level 1

Simplify the following expression:

sin ⁡ θ cos ⁡ θ + cos ⁡ θ 1 + sin ⁡ θ \large \frac{\sin\theta }{\cos\theta }+\frac{\cos\theta }{1+\sin\theta }

This is a part of a set: It's trigo time

sec ⁡ θ \sec\theta cot ⁡ θ \cot\theta cos ⁡ 2 θ \cos^2\theta tan ⁡ θ \tan\theta sin ⁡ 2 θ \sin^2\theta

View wiki
Syed Hamza Khalid by Syed Hamza Khalid , 17, France

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

4 solutions

Syed Hamza Khalid
May 12, 2017

**Let θ \theta be x

5 Helpful 0 Interesting 0 Brilliant 0 Confused
Chew-Seong Cheong
May 13, 2017

sin ⁡ θ cos ⁡ θ + cos ⁡ θ 1 + sin ⁡ θ = sin ⁡ θ + sin ⁡ 2 θ + cos ⁡ 2 θ cos ⁡ θ ( 1 + sin ⁡ θ ) Note that sin ⁡ 2 x + cos ⁡ 2 x = 1 = sin ⁡ θ + 1 cos ⁡ θ ( 1 + sin ⁡ θ ) = 1 cos ⁡ θ = sec ⁡ θ \begin{aligned} \frac {\sin \theta}{\cos \theta} + \frac {\cos \theta}{1+\sin \theta} & = \frac {\sin \theta + \color{#3D99F6} \sin^2 \theta + \cos^2 \theta}{\cos \theta(1+\sin \theta)} & \small \color{#3D99F6} \text{Note that }\sin^2 x + \cos^2 x = 1 \\ & = \frac {\sin \theta + \color{#3D99F6} 1}{\cos \theta(1+\sin \theta)} \\ & = \frac 1{\cos \theta} \\ & = \sec \theta \end{aligned}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

You can possibly try out some of my hard trigonometry questions like:

https://brilliant.org/problems/its-trigo-time-7/

Syed Hamza Khalid - 4 years ago
Zach Abueg
May 12, 2017

sin ⁡ θ cos ⁡ θ + cos ⁡ θ 1 + sin ⁡ θ \displaystyle \frac {\sin \theta}{\cos \theta} + \frac {\cos \theta}{1 + \sin \theta}

= tan ⁡ θ + cos ⁡ θ ⋅ ( 1 − sin ⁡ θ ) ( 1 + sin ⁡ θ ) ⋅ ( 1 − sin ⁡ θ ) \displaystyle = \tan \theta + \frac {\cos \theta \cdot (1 - \sin \theta)}{(1 + \sin \theta) \cdot (1 - \sin \theta)}

= tan ⁡ θ + cos ⁡ θ − cos ⁡ θ sin ⁡ θ cos ⁡ 2 θ \displaystyle = \tan \theta + \frac {\cos \theta - \cos \theta \sin \theta}{\cos^2 \theta}

= tan ⁡ θ + sec ⁡ θ − tan ⁡ θ \displaystyle = \tan \theta + \sec \theta - \tan \theta

= sec ⁡ θ \displaystyle = \sec \theta

3 Helpful 0 Interesting 0 Brilliant 0 Confused

sin ⁡ θ cos ⁡ θ + cos ⁡ θ 1 + sin ⁡ θ \dfrac{\sin \theta}{\cos \theta}+\dfrac{\cos \theta}{1+\sin \theta}

⟹ \implies the L C D LCD is cos ⁡ θ ( 1 + sin ⁡ θ ) \cos \theta(1+\sin \theta)

= sin ⁡ θ ( 1 + sin ⁡ θ ) + cos ⁡ θ ( cos ⁡ θ ) cos ⁡ θ ( 1 + sin ⁡ θ ) =\dfrac{\sin \theta(1+\sin \theta)+\cos \theta(\cos \theta)}{\cos \theta(1+\sin \theta)}

= sin ⁡ θ + sin ⁡ 2 θ + c o s 2 θ cos ⁡ θ ( 1 + sin ⁡ θ ) =\dfrac{\sin \theta + \sin^2 \theta + cos^2 \theta}{\cos \theta(1+\sin \theta)}

= sin ⁡ θ + 1 cos ⁡ θ ( 1 + sin ⁡ θ ) =\dfrac{\sin \theta + 1}{\cos \theta(1+\sin \theta)}

= 1 cos ⁡ θ =\dfrac{1}{\cos \theta}

= sec ⁡ θ =\sec \theta

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...