Question

If $ \sin \theta + \cos \theta = 0 $ and $ 0 < \theta < \pi $ then $ \theta $ is equal to
 $ A)0 $
 $ B)\dfrac{\pi }{4} $
 $ C)\dfrac{\pi }{2} $
 $ D)\dfrac{{3\pi }}{4} $

Answer 1

Hint: First, we need to analyze the given information which is in the trigonometric form.
The trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.
 From the given, we asked to calculate the value of $ \theta $ when $ \sin \theta + \cos \theta = 0 $ and also range of the function is given as $ 0 < \theta < \pi $ , so we need to know the formulas in sine, cos, tangent in the trigonometry.
Formula used:
 $ \dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta $
 $ {\tan ^{ - 1}}( - 1) = \dfrac{{3\pi }}{4} $

Complete step by step answer:
Since from the given that we have to trigonometric functions as $ \sin \theta + \cos \theta = 0 $
Now subtract $ \cos \theta $ in both sides of the function given above, then we get $ \sin \theta + \cos \theta - \cos \theta = - \cos \theta $
By the subtraction operation, the common values get cancel each other; thus, we get $ \sin \theta = - \cos \theta $
Since the right-hand value can be rewritten as in the form of $ - 1 \times \cos \theta $ and substituting this value in the above we get $ \sin \theta = - \cos \theta \Rightarrow \sin \theta = - 1 \times \cos \theta $
Now divide both the right and left-hand side values with the trigonometric value $ \cos \theta $ then we get $ \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{ - 1 \times \cos \theta }}{{\cos \theta }} $ and further canceled the common values we have $ \dfrac{{\sin \theta }}{{\cos \theta }} = - 1 $
We know that $ \dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta $ substituting in the above value we get $ \dfrac{{\sin \theta }}{{\cos \theta }} = - 1 \Rightarrow \tan \theta = - 1 $
Since the condition of the value $ \theta $ is $ 0 < \theta < \pi $ (strictly less than zero and strictly greater than $ \pi $ )
Bu using the inverse of trigonometric values, which is $ \tan \theta = x $ can be written in the form of $ x = {\tan ^{ - 1}}\theta $
Applying this condition, we get $ \tan \theta = - 1 \Rightarrow \theta = {\tan ^{ - 1}}( - 1) $
Since the range of $ 0 < \theta < \pi $ and using the quadrant table we have the negative sign value of tangent with minus one.
Hence, we get $ \theta = {\tan ^{ - 1}}( - 1) \Rightarrow \dfrac{{3\pi }}{4} $

So, the correct answer is “Option D”.

Note: Simply using the trigonometric value of sine and cos for the sec and cosec we solved the given function.
Also, the value of the $ \theta = {\tan ^{ - 1}}( - 1) $ is generally $ \theta = {\tan ^{ - 1}}( - 1) \Rightarrow \dfrac{{ - \pi }}{4} $ but which is not in the given range of $ 0 < \theta < \pi $ . So, we converted the tangent in the second quadrant to get the resultant.
In total there are six trigonometric values which are sine, cos, tan, sec, cosec, cot while all the values have been relation like $ \dfrac{{\sin }}{{\cos }} = \tan $ and $ \tan = \dfrac{1}{{\cot }} $