# Trigonometry Identity: $\tan \theta\sin \theta + \cos \theta = \sec \theta$

Sorry if my question seems too simple. I cannot find a proof and my text book does not provide one either. I am supposed to prove:

$$\tan \theta \times \sin \theta + \cos \theta = \sec \theta$$

I know that $\sec = \frac{1}{\cos\theta}$. But I do not know how to prove that $\tan \theta \times \sin \theta + \cos \theta = \frac{1}{\cos \theta}$.

I appreciate if someone point me to the right direction.

• First write $\tan\theta\sin\theta+\cos\theta={\sin^2\theta\over\cos\theta}+{\cos^2\theta \over\cos\theta}$. – David Mitra Aug 18 '14 at 11:03
• I would start by writing the whole equation out in terms of sine and cosine functions. – Mark Bennet Aug 18 '14 at 11:03
• Use the identity $sin^2 x + cos^2 x =1$. – cejvan Aug 18 '14 at 11:04
• Is your problem solved? If it is, you should accept an answer or provide your own. – AlexR Sep 10 '14 at 11:55

\begin{align*} \tan \theta \sin \theta + \cos \theta & \stackrel{\text{def.}}= \frac{\sin^2 \theta}{\cos \theta} + \cos \theta \\ & \stackrel{\text{Pythagoras}}= \frac{1-\cos^2 \theta}{\cos \theta} + \cos \theta \\ & = \frac1{\cos\theta} - \cos \theta + \cos \theta \\ & \stackrel{\text{def.}}= \sec\theta \end{align*} Where we use the definitions of $\tan \theta$ and $\sec\theta$ plus Pythagoras' theorem $\sin^2 \theta + \cos^2 \theta = 1$.

To Prove :$$\tan \theta\sin \theta + \cos \theta = \sec \theta$$

L.H.S :

$\tan \theta\sin \theta + \cos \theta$ $\implies \large\frac{\sin \theta}{\cos \theta}\times{\sin \theta} + \cos \theta$

$\implies \large \frac{\sin^2 \theta +\cos^2 \theta}{\cos \theta}$ $\implies \large \frac{1}{\cos \theta}$

$\implies \large {\sec \theta}$

:)

• $\sin \theta \times \sin \theta + cos \theta$ should be equal to $\sin^{2}\theta + \cos\theta$. Why you square $\cos \theta$ and wrote $\sin^{2}\theta + cos^{2} \theta$? – bman Aug 18 '14 at 11:37
• There is a $\cos \theta$ in Denominator, cross multiply with $\cos \theta$ gives you $cos^2 \theta$. Can you see it/ – MonK Aug 18 '14 at 12:15

AS mentioned in a comment above $$\tan \theta =\dfrac{\sin \theta}{\cos \theta}.$$

$$\dfrac{\sin \theta}{\cos \theta}.\sin \theta + \cos \theta$$
from this can you add the terms together? and use $\sin^2 \theta +\cos^2 \theta = 1$
First note that $\tan\theta =\frac{\sin\theta}{\cos\theta}$ $\sin^{2}\theta+\cos^{2}\theta=1$ $\frac{a}{b}\pm c=\frac{a\pm bc}{b}$ So then $\tan\theta\sin\theta+\cos\theta= \frac{\sin^{2}\theta}{\cos\theta}+\cos\theta= \frac{\sin^{2}\theta+\cos^{2}\theta}{\cos\theta}=\frac{1}{\cos\theta}=\sec\theta$