Question

Sin theta - cos theta=7/13 find sin theta + cos theta

Answer 1

Answer is in the attachment..

Hope it helps :)

Answer 2

The value of $\sin \theta +\cos \theta=\dfrac{17}{13}$ .

Explanation:

Given : $\sin \theta -\cos \theta= \dfrac{7}{13}$

To find : $\sin \theta +\cos \theta$

Consider :   $\sin \theta -\cos \theta= \dfrac{7}{13}$

Squaring both sides , we get

$(\sin \theta -\cos \theta)^2= (\dfrac{7}{13})^2$

$\Rightarrow\ \sin^2 \theta +\cos^2 \theta-2\sin \theta\cos \theta= \dfrac{49}{169}$

$\Rightarrow\ 1-2\sin \theta\cos \theta= \dfrac{49}{169}$   [Since , $\sin^2 x +\cos^2x=1$]

$\Rightarrow\ 2\sin \theta\cos \theta= 1-\dfrac{49}{169}=\dfrac{169-49}{169}=\dfrac{120}{169}$   (1)

Consider $(\sin \theta +\cos \theta)^2= \sin^2 \theta +\cos^2 \theta+2\sin \theta\cos \theta$

From (1),

$(\sin \theta +\cos \theta)^2= 1+\dfrac{120}{169}=\dfrac{169+120}{169}$

i.e. $(\sin \theta +\cos \theta)^2=\dfrac{289}{169}$

Taking square root on both sides , we get

. $\sin \theta +\cos \theta=\sqrt{\dfrac{289}{169}}=\dfrac{17}{13}$

Hence, the value of $\sin \theta +\cos \theta=\dfrac{17}{13}$

# Learn more :

If Sin theta

-Cos theta=½find Sin theta+Cos theta

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