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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.

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    $\begingroup$ First write $\tan\theta\sin\theta+\cos\theta={\sin^2\theta\over\cos\theta}+{\cos^2\theta \over\cos\theta}$. $\endgroup$ – David Mitra Aug 18 '14 at 11:03
  • $\begingroup$ I would start by writing the whole equation out in terms of sine and cosine functions. $\endgroup$ – Mark Bennet Aug 18 '14 at 11:03
  • $\begingroup$ Use the identity $sin^2 x + cos^2 x =1$. $\endgroup$ – cejvan Aug 18 '14 at 11:04
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    $\begingroup$ Is your problem solved? If it is, you should accept an answer or provide your own. $\endgroup$ – AlexR Sep 10 '14 at 11:55
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$$\begin{align*} \tan \theta \sin \theta + \cos \theta & \stackrel{\text{def.}}= \frac{\sin^2 \theta}{\cos \theta} + \cos \theta \\ & \stackrel{\text{Pythagoras}}= \frac{1-\cos^2 \theta}{\cos \theta} + \cos \theta \\ & = \frac1{\cos\theta} - \cos \theta + \cos \theta \\ & \stackrel{\text{def.}}= \sec\theta \end{align*}$$ Where we use the definitions of $\tan \theta$ and $\sec\theta$ plus Pythagoras' theorem $\sin^2 \theta + \cos^2 \theta = 1$.

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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} $

:)

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

then your L.H.S becomes

$$ \dfrac{\sin \theta}{\cos \theta}.\sin \theta + \cos \theta $$

from this can you add the terms together? and use $\sin^2 \theta +\cos^2 \theta = 1$

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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 \[ \tan\theta\sin\theta+\cos\theta= \frac{\sin^{2}\theta}{\cos\theta}+\cos\theta= \frac{\sin^{2}\theta+\cos^{2}\theta}{\cos\theta}=\frac{1}{\cos\theta}=\sec\theta \]

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