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Question: If \[\sin \theta -\cos \theta =0\]then the value is \[({{\sin }^{4}}\theta +{{\cos }^{4}}\theta )\] ...

If sin⁡θ−cos⁡θ=0\sin \theta -\cos \theta =0then the value is (sin⁡4θ+cos⁡4θ)({{\sin }^{4}}\theta +{{\cos }^{4}}\theta )
A. 1
B. 34\dfrac{3}{4}
C. 12\dfrac{1}{2}
D. 12\dfrac{1}{2}

Explanation

Solution

Hint: We can solve this question by finding the value of θ\theta from the given condition and then we can find the value of the required expression at a particular value of θ\theta .
Complete step-by-step answer:
sin⁡θ−cos⁡θ=0\sin \theta -\cos \theta =0
sin⁡θ=cos⁡θ\sin \theta =\cos \theta ...........................................(i)
Now we can use cos⁡(θ)=sin⁡(90∘−θ)\cos (\theta )=\sin \left( {{90}^{\circ }}-\theta \right)
So we can write equation (i) as
sin⁡θ=sin⁡(90∘−θ)\sin \theta =\sin ({{90}^{\circ }}-\theta )
On comparing
⇒θ=90∘−θ\Rightarrow \theta ={{90}^{\circ }}-\theta
⇒θ+θ=90∘\Rightarrow \theta +\theta ={{90}^{\circ }}
⇒2θ=90∘\Rightarrow 2\theta ={{90}^{\circ }}
⇒θ=90∘2\Rightarrow \theta =\dfrac{{{90}^{\circ }}}{2}
⇒θ=45∘\Rightarrow \theta ={{45}^{\circ }}
Given expression is
⇒sin⁡4θ+cos⁡4θ\Rightarrow {{\sin }^{4}}\theta +{{\cos }^{4}}\theta
At θ=45∘\theta ={{45}^{\circ }}
⇒(sin⁡45∘)4+(cos⁡45∘)4\Rightarrow {{\left( \sin {{45}^{\circ }} \right)}^{4}}+{{\left( \cos {{45}^{\circ }} \right)}^{4}}
⇒(12)4+(12)4\Rightarrow {{\left( \dfrac{1}{\sqrt{2}} \right)}^{4}}+{{\left( \dfrac{1}{\sqrt{2}} \right)}^{4}} \left\\{ \because \sin 45{}^\circ =\dfrac{1}{\sqrt{2}},\cos 45{}^\circ =\dfrac{1}{\sqrt{2}} \right\\}
⇒14+14\Rightarrow \dfrac{1}{4}+\dfrac{1}{4}
⇒12\Rightarrow \dfrac{1}{2}
Hence option C is correct.
Note: In this type of question we need to be careful about choosing the value of unknown angles. In the given question there is no range of θ\theta given. So we choose an angle in the first quadrant. But if there is a range we need to choose the value of angle according to that range.
We can solve this question by using tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan (\theta )=\dfrac{\sin (\theta )}{\cos (\theta )}
⇒sin⁡θ−cos⁡θ=0\Rightarrow \sin \theta -\cos \theta =0
⇒sin⁡θ=cos⁡θ\Rightarrow \sin \theta =\cos \theta
⇒sin⁡θcos⁡θ=1\Rightarrow \dfrac{\sin \theta }{\cos \theta }=1
⇒tan⁡θ=1\Rightarrow \tan \theta =1
.⇒tan⁡θ=tan⁡45∘\Rightarrow \tan \theta =\tan 45{}^\circ
⇒θ=45∘\Rightarrow \theta =45{}^\circ
Now we can substitute the value of θ\theta and calculate the value of the given expression.