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Question: If \[f(\theta ) = \dfrac{{1 - \sin 2\theta + \cos 2\theta }}{{2\cos 2\theta }}\]then the value of \[...

If f(θ)=1−sin⁡2θ+cos⁡2θ2cos⁡2θf(\theta ) = \dfrac{{1 - \sin 2\theta + \cos 2\theta }}{{2\cos 2\theta }}then the value of f(11∘).f(34∘)f\left( {11^\circ } \right).f\left( {34^\circ } \right)
A.12\dfrac{1}{2}
B.34\dfrac{3}{4}
C.14\dfrac{1}{4}
D.11

Explanation

Solution

We can’t directly comprehend the value of the expression as the given angles, 11∘11^\circ or 34∘34^\circ are not in the standard trigonometric value. But what we can see is that the sum of these two angles is 45∘45^\circ which perhaps is in the standard trigonometric values. In order to simplify the given expression, we must expand the sines and cosines into values containing a simple θ\theta and not a 2θ2\theta , which can be achieved by using the formulae for sin⁡2θ\sin 2\theta and cos⁡2θ\cos 2\theta .

Formula Used:
In the given question, we shall be manipulating the following values in the expression of f(θ)f\left( \theta \right) and applying formulae to them:
1=sin⁡2θ+cos⁡2θ...(i)1 = {\sin ^2}\theta + {\cos ^2}\theta {\rm{ }}...\left( i \right)
sin⁡2θ=sin⁡(θ+θ)\sin 2\theta = \sin \left( {\theta + \theta } \right)
We know, sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B
So, sin⁡2θ=sin⁡θcos⁡θ+cos⁡θsin⁡θ=2sin⁡θcos⁡θ...(ii)\sin 2\theta = \sin \theta \cos \theta + \cos \theta \sin \theta = 2\sin \theta \cos \theta {\rm{ }}...\left( {ii} \right)
Also, cos⁡2θ=cos⁡(θ+θ)\cos 2\theta = \cos \left( {\theta + \theta } \right)
We know, cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B
So, cos⁡2θ=cos⁡θ.cos⁡θ−sin⁡θ.sin⁡θ=cos⁡2θ−sin⁡2θ...(iii)\cos 2\theta = \cos \theta .\cos \theta - \sin \theta .\sin \theta = {\cos ^2}\theta - {\sin ^2}\theta {\rm{ }}...\left( {iii} \right)
We shall also be using the difference formula for tangent, which is
tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡A×tan⁡B...(iv)\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A \times \tan B}}{\rm{ }}...\left( {iv} \right)

Complete step-by-step answer:
f(θ)=1−sin⁡2θ+cos⁡2θ2cos⁡2θf(\theta ) = \dfrac{{1 - \sin 2\theta + \cos 2\theta }}{{2\cos 2\theta }}
Putting in the values from (i), (ii), and (iii) into f(θ)f\left( \theta \right), we have
f(θ)=(sin⁡2θ+cos⁡2θ)−(2sin⁡θcos⁡θ)+(cos⁡2θ−sin⁡2θ)2(cos⁡2θ−sin⁡2θ)f\left( \theta \right) = \dfrac{{\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) - \left( {2\sin \theta \cos \theta } \right) + \left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right)}}{{2\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right)}}
f(θ)=2cos⁡2θ−2sin⁡θcos⁡θ2(cos⁡2θ−sin⁡2θ)f\left( \theta \right) = \dfrac{{2{{\cos }^2}\theta - 2\sin \theta \cos \theta }}{{2\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right)}}
In the numerator, from the two expressions we can common out 2cos⁡θ2\cos \theta and from the denominator we can make out an elementary algebraic expression of difference of squares: a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)
f(θ)=2cos⁡θ(cos⁡θ−sin⁡θ)2(cos⁡θ−sin⁡θ)(cos⁡θ+sin⁡θ)f\left( \theta \right) = \dfrac{{2\cos \theta \left( {\cos \theta - \sin \theta } \right)}}{{2\left( {\cos \theta - \sin \theta } \right)\left( {\cos \theta + \sin \theta } \right)}}
Now clearly, in both the numerator and denominator we have a common element 2(cos⁡θ−sin⁡θ)2\left( {\cos \theta - \sin \theta } \right), which we can cancel out to reduce the given expression’s fraction, so we have got,
f(θ)=cos⁡θcos⁡θ+sin⁡θf\left( \theta \right) = \dfrac{{\cos \theta }}{{\cos \theta + \sin \theta }}
Now we have to take cos⁡θ\cos \theta common from both the numerator and denominator and cancel it out to again reduce the expression’s fraction, so we have got,
f(θ)=cos⁡θcos⁡θ(1+sin⁡θcos⁡θ)f\left( \theta \right) = \dfrac{{\cos \theta }}{{\cos \theta \left( {1 + \dfrac{{\sin \theta }}{{\cos \theta }}} \right)}}
Also, we know that sin⁡θcos⁡θ=tan⁡θ\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta
So, f(θ)=11+tan⁡θf\left( \theta \right) = \dfrac{1}{{1 + \tan \theta }}
Since now we have simplified the expression into a single function (tangent), we can put in the values and then work from there, so we have got,
f(11∘)×f(34∘)f\left( {11^\circ } \right) \times f\left( {34^\circ } \right)
=11+tan⁡11∘×11+tan⁡34∘= \dfrac{1}{{1 + \tan 11^\circ }} \times \dfrac{1}{{1 + \tan 34^\circ }}
=11+tan⁡11∘×11+tan⁡(45−11)∘= \dfrac{1}{{1 + \tan 11^\circ }} \times \dfrac{1}{{1 + \tan \left( {45 - 11} \right)^\circ }}
Now, using the formula from (iv), we have
=11+tan⁡11∘×11+tan⁡45∘−tan⁡11∘1+tan⁡45∘tan⁡11∘= \dfrac{1}{{1 + \tan 11^\circ }} \times \dfrac{1}{{1 + \dfrac{{\tan 45^\circ - \tan 11^\circ }}{{1 + \tan 45^\circ \tan 11^\circ }}}}
Now we can substitute the value of tan⁡45∘=1\tan 45^\circ = 1 in the expression,
=11+tan⁡11∘×11+1−tan⁡11∘1+1×tan⁡11∘= \dfrac{1}{{1 + \tan 11^\circ }} \times \dfrac{1}{{1 + \dfrac{{1 - \tan 11^\circ }}{{1 + 1 \times \tan 11^\circ }}}}
By taking the L.C.M we get,
=11+tan⁡11∘×11+tan⁡11∘1+tan⁡11∘+1−tan⁡11∘1+tan⁡11∘= \dfrac{1}{{1 + \tan 11^\circ }} \times \dfrac{1}{{\dfrac{{1 + \tan 11^\circ }}{{1 + \tan 11^\circ }} + \dfrac{{1 - \tan 11^\circ }}{{1 + \tan 11^\circ }}}}
=11+tan⁡11∘×11+tan⁡11∘+1−tan⁡11∘1+tan⁡11∘= \dfrac{1}{{1 + \tan 11^\circ }} \times \dfrac{1}{{\dfrac{{1 + \tan 11^\circ + 1 - \tan 11^\circ }}{{1 + \tan 11^\circ }}}}
=11+tan⁡11∘×121+tan⁡11∘= \dfrac{1}{{1 + \tan 11^\circ }} \times \dfrac{1}{{\dfrac{2}{{1 + \tan 11^\circ }}}}
=11+tan⁡11∘×1+tan⁡11∘2=12= \dfrac{1}{{1 + \tan 11^\circ }} \times \dfrac{{1 + \tan 11^\circ }}{2} = \dfrac{1}{2}

So, the answer to the given question is A) 12\dfrac{1}{2}.

Note: In order to solve these questions effectively, it’s best to first identify if the sum/difference of the angles in the problem is something which is in the standard trigonometric values and then shape either of the two angles in form of that value. At first these problems might look intimidating but the answer lies in reducing them with the help of basic trigonometric formulae.