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Question: If \( \sin \theta + \cos \theta = 0 \) and \( 0 < \theta < \pi \) then \( \theta \) is equal to ...

If sin⁡θ+cos⁡θ=0\sin \theta + \cos \theta = 0 and 0<θ<π0 < \theta < \pi then θ\theta is equal to
A)0A)0
B)π4B)\dfrac{\pi }{4}
C)π2C)\dfrac{\pi }{2}
D)3π4D)\dfrac{{3\pi }}{4}

Explanation

Solution

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 θ\theta when sin⁡θ+cos⁡θ=0\sin \theta + \cos \theta = 0 and also range of the function is given as 0<θ<π0 < \theta < \pi , so we need to know the formulas in sine, cos, tangent in the trigonometry.
Formula used:
sin⁡θcos⁡θ=tan⁡θ\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta
tan⁡−1(−1)=3π4{\tan ^{ - 1}}( - 1) = \dfrac{{3\pi }}{4}

Complete step by step answer:
Since from the given that we have to trigonometric functions as sin⁡θ+cos⁡θ=0\sin \theta + \cos \theta = 0
Now subtract cos⁡θ\cos \theta in both sides of the function given above, then we get sin⁡θ+cos⁡θ−cos⁡θ=−cos⁡θ\sin \theta + \cos \theta - \cos \theta = - \cos \theta
By the subtraction operation, the common values get cancel each other; thus, we get sin⁡θ=−cos⁡θ\sin \theta = - \cos \theta
Since the right-hand value can be rewritten as in the form of −1×cos⁡θ- 1 \times \cos \theta and substituting this value in the above we get sin⁡θ=−cos⁡θ⇒sin⁡θ=−1×cos⁡θ\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⁡θ\cos \theta then we get sin⁡θcos⁡θ=−1×cos⁡θcos⁡θ\dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{ - 1 \times \cos \theta }}{{\cos \theta }} and further canceled the common values we have sin⁡θcos⁡θ=−1\dfrac{{\sin \theta }}{{\cos \theta }} = - 1
We know that sin⁡θcos⁡θ=tan⁡θ\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta substituting in the above value we get sin⁡θcos⁡θ=−1⇒tan⁡θ=−1\dfrac{{\sin \theta }}{{\cos \theta }} = - 1 \Rightarrow \tan \theta = - 1
Since the condition of the value θ\theta is 0<θ<π0 < \theta < \pi (strictly less than zero and strictly greater than π\pi )
Bu using the inverse of trigonometric values, which is tan⁡θ=x\tan \theta = x can be written in the form of x=tan⁡−1θx = {\tan ^{ - 1}}\theta
Applying this condition, we get tan⁡θ=−1⇒θ=tan⁡−1(−1)\tan \theta = - 1 \Rightarrow \theta = {\tan ^{ - 1}}( - 1)
Since the range of 0<θ<π0 < \theta < \pi and using the quadrant table we have the negative sign value of tangent with minus one.
Hence, we get θ=tan⁡−1(−1)⇒3π4\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 θ=tan⁡−1(−1)\theta = {\tan ^{ - 1}}( - 1) is generally θ=tan⁡−1(−1)⇒−π4\theta = {\tan ^{ - 1}}( - 1) \Rightarrow \dfrac{{ - \pi }}{4} but which is not in the given range of 0<θ<π0 < \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 sin⁡cos⁡=tan⁡\dfrac{{\sin }}{{\cos }} = \tan and tan⁡=1cot⁡\tan = \dfrac{1}{{\cot }}