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Question: If \[\sin \theta = \dfrac{{\sqrt 3 }}{2}\]and\[\cos \phi = \dfrac{1}{{\sqrt 2 }}\] Find the value of...

If sin⁡θ=32\sin \theta = \dfrac{{\sqrt 3 }}{2}andcos⁡ϕ=12\cos \phi = \dfrac{1}{{\sqrt 2 }} Find the value of tan⁡θ−tan⁡ϕ1+tan⁡θtan⁡ϕ\dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }}

Explanation

Solution

Here we use the table which gives us the value of trigonometric functions at different angles.
With the help of the table we find the values of θ\theta and ϕ\phi . Substitute the values of obtained angles in the numerator and denominator and write the values of trigonometric functions using the table.

  • Table for trigonometric functions like sine, cosine and tan at angles 0∘,30∘,45∘,60∘,90∘{0^ \circ },{30^ \circ },{45^ \circ },{60^ \circ },{90^ \circ } is
ANGLEFUNCTION0∘{0^ \circ }30∘{30^ \circ }45∘{45^ \circ }60∘{60^ \circ }90∘{90^ \circ }
Sin012\dfrac{1}{2}12\dfrac{1}{{\sqrt 2 }}32\dfrac{{\sqrt 3 }}{2}1
Cos132\dfrac{{\sqrt 3 }}{2}12\dfrac{1}{{\sqrt 2 }}12\dfrac{1}{2}0
Tan013\dfrac{1}{{\sqrt 3 }}13\sqrt 3 Not defined

Complete step-by-step answer:
We are given that sin⁡θ=32\sin \theta = \dfrac{{\sqrt 3 }}{2}
So, here the angle is θ\theta
We know that the value of sine function is 32\dfrac{{\sqrt 3 }}{2}when the angle is 60∘{60^ \circ }
Therefore, we can write value of 32=sin⁡60∘\dfrac{{\sqrt 3 }}{2} = \sin {60^ \circ }
⇒sin⁡θ=sin⁡60∘\Rightarrow \sin \theta = \sin {60^ \circ }
We know the value of trigonometric function will be equal if the angles will be of equal measure
⇒θ=60∘\Rightarrow \theta = {60^ \circ } … (1)
Also,cos⁡ϕ=12\cos \phi = \dfrac{1}{{\sqrt 2 }}
So, here the angle is ϕ\phi
We know that the value of cosine function is 12\dfrac{1}{{\sqrt 2 }}when the angle is 45∘{45^ \circ }
Therefore, we can write value of 12=cos⁡45∘\dfrac{1}{{\sqrt 2 }} = \cos {45^ \circ }
⇒cos⁡ϕ=cos⁡45∘\Rightarrow \cos \phi = \cos {45^ \circ }
We know the value of trigonometric function will be equal if the angles will be of equal measure
⇒ϕ=45∘\Rightarrow \phi = {45^ \circ } … (2)
Now we have to find the value of tan⁡θ−tan⁡ϕ1+tan⁡θtan⁡ϕ\dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }}
Substitute the values of angle θ\theta from equation (1) and angle ϕ\phi from equation (2) in numerator and denominator of tan⁡θ−tan⁡ϕ1+tan⁡θtan⁡ϕ\dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }}
⇒tan⁡θ−tan⁡ϕ1+tan⁡θtan⁡ϕ=tan⁡60∘−tan⁡45∘1+tan⁡60∘tan⁡45∘\Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = \dfrac{{\tan {{60}^ \circ } - \tan {{45}^ \circ }}}{{1 + \tan {{60}^ \circ }\tan {{45}^ \circ }}} … (3)
Now we find the values of tan⁡60∘\tan {60^ \circ }andtan⁡45∘\tan {45^ \circ }from the table.
tan⁡60∘=3\tan {60^ \circ } = \sqrt 3 andtan⁡45∘=1\tan {45^ \circ } = 1
Substitute the values in numerator and denominator of equation (3)
⇒tan⁡θ−tan⁡ϕ1+tan⁡θtan⁡ϕ=3−11+3×1\Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = \dfrac{{\sqrt 3 - 1}}{{1 + \sqrt 3 \times 1}}
⇒tan⁡θ−tan⁡ϕ1+tan⁡θtan⁡ϕ=3−13+1\Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}
Rationalize the term in RHS by multiplying both numerator and denominator by 3−1\sqrt 3 - 1
⇒tan⁡θ−tan⁡ϕ1+tan⁡θtan⁡ϕ=3−13+1×3−13−1\Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} \times \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}
Multiply the numerator with numerator and denominator with denominator
⇒tan⁡θ−tan⁡ϕ1+tan⁡θtan⁡ϕ=(3−1)(3−1)(3+1)(3−1)\Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = \dfrac{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}}
⇒tan⁡θ−tan⁡ϕ1+tan⁡θtan⁡ϕ=(3−1)2(3+1)(3−1)\Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = \dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}}
Use the property(a−b)2=a2+b2−2ab{(a - b)^2} = {a^2} + {b^2} - 2abin numerator and the property (a+b)(a−b)=a2−b2(a + b)(a - b) = {a^2} - {b^2}in denominator of RHS
⇒tan⁡θ−tan⁡ϕ1+tan⁡θtan⁡ϕ=((3)2+(1)2−2×(3×1))(3)2−(1)2\Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = \dfrac{{\left( {{{(\sqrt 3 )}^2} + {{(1)}^2} - 2 \times (\sqrt 3 \times 1)} \right)}}{{{{(\sqrt 3 )}^2} - {{(1)}^2}}}
⇒tan⁡θ−tan⁡ϕ1+tan⁡θtan⁡ϕ=3+1−233−1\Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = \dfrac{{3 + 1 - 2\sqrt 3 }}{{3 - 1}}
⇒tan⁡θ−tan⁡ϕ1+tan⁡θtan⁡ϕ=4−232\Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = \dfrac{{4 - 2\sqrt 3 }}{2}
Take 2 common from numerator in RHS of the equation
⇒tan⁡θ−tan⁡ϕ1+tan⁡θtan⁡ϕ=2(2−3)2\Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = \dfrac{{2(2 - \sqrt 3 )}}{2}
Cancel the same factors from numerator and denominator.
⇒tan⁡θ−tan⁡ϕ1+tan⁡θtan⁡ϕ=2−3\Rightarrow \dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} = 2 - \sqrt 3

So, the value of tan⁡θ−tan⁡ϕ1+tan⁡θtan⁡ϕ\dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }} is 2−32 - \sqrt 3 .

Note: Students might try to use the property of tan⁡A−tan⁡B=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan A - \tan B = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}} to write tan⁡θ−tan⁡ϕ1+tan⁡θtan⁡ϕ\dfrac{{\tan \theta - \tan \phi }}{{1 + \tan \theta \tan \phi }}. But then we will have the term equal to tan⁡(θ−ϕ)=tan⁡15∘\tan (\theta - \phi ) = \tan {15^ \circ }. Since, we don’t know the value of tan⁡15∘\tan {15^ \circ } we avoid this process. Also, many students leave the answer with an irrational number i.e. under root value in the denominator, keep in mind the answer should always be rationalized and in lowest form.