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Question: If \[\theta \] is an acute angle and \[\sin \theta = \cos \theta \], find the value of \[2{\tan ^2}\...

If θ\theta is an acute angle and sin⁡θ=cos⁡θ\sin \theta = \cos \theta , find the value of 2tan⁡2θ+sin⁡2θ−12{\tan ^2}\theta + {\sin ^2}\theta - 1.

Explanation

Solution

We use the given equality between sine and cosine of an angle. Divide both sides of the equality with such a trigonometric function so we can make out the value of angle using tangent of the angle. Find the angle and substitute it in the equation whose value we have to find. Check whether the angle is acute or not. Use a table of trigonometric values to substitute values in the equation.

  • Acute angle: Any angle having measure less than 90∘{90^ \circ } is called an acute angle.
  • tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{{\sin x}}{{\cos x}}

Complete step-by-step solution:
We are given sin⁡θ=cos⁡θ\sin \theta = \cos \theta ....................… (1)
We divide both sides of equation (1) by cos⁡θ\cos \theta
⇒sin⁡θcos⁡θ=cos⁡θcos⁡θ\Rightarrow \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{\cos \theta }}{{\cos \theta }}
Cancel same terms in RHS of the equation and use tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{{\sin x}}{{\cos x}}in LHS of the equation
⇒tan⁡θ=1\Rightarrow \tan \theta = 1........................… (2)
Now we know from the table of trigonometric functions thattan⁡45∘=1\tan {45^ \circ } = 1. Substitute the value of 1 in equation (2)
⇒tan⁡θ=tan⁡45∘\Rightarrow \tan \theta = \tan {45^ \circ }
Take inverse tangent function on both sides of the equation
⇒tan⁡−1(tan⁡θ)=tan⁡−1(tan⁡45∘)\Rightarrow {\tan ^{ - 1}}\left( {\tan \theta } \right) = {\tan ^{ - 1}}\left( {\tan {{45}^ \circ }} \right)
Since we know f−1(f(x))=x{f^{ - 1}}(f(x)) = x
⇒θ=45∘\Rightarrow \theta = {45^ \circ }.........................… (3)
Since, θ=45∘<90∘\theta = {45^ \circ } < {90^ \circ }, so the angle is an acute angle.
Now substitute the value from equation (3) in the equation 2tan⁡2θ+sin⁡2θ−12{\tan ^2}\theta + {\sin ^2}\theta - 1
⇒2tan⁡2θ+sin⁡2θ−1=2tan⁡2(45∘)+sin⁡2(45∘)−1\Rightarrow 2{\tan ^2}\theta + {\sin ^2}\theta - 1 = 2{\tan ^2}({45^ \circ }) + {\sin ^2}({45^ \circ }) - 1
⇒2tan⁡2θ+sin⁡2θ−1=2[tan⁡(45∘)]2+[sin⁡(45∘)]2−1\Rightarrow 2{\tan ^2}\theta + {\sin ^2}\theta - 1 = 2{\left[ {\tan ({{45}^ \circ })} \right]^2} + {\left[ {\sin ({{45}^ \circ })} \right]^2} - 1.................… (4)
Substitute the value of tan⁡45∘=1,sin⁡45∘=12\tan {45^ \circ } = 1,\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }} in equation (4)
⇒2tan⁡2θ+sin⁡2θ−1=2(1)2+(12)2−1\Rightarrow 2{\tan ^2}\theta + {\sin ^2}\theta - 1 = 2{(1)^2} + {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2} - 1
⇒2tan⁡2θ+sin⁡2θ−1=2+12−1\Rightarrow 2{\tan ^2}\theta + {\sin ^2}\theta - 1 = 2 + \dfrac{1}{2} - 1
Take LCM in RHS of the equation
⇒2tan⁡2θ+sin⁡2θ−1=4+1−22\Rightarrow 2{\tan ^2}\theta + {\sin ^2}\theta - 1 = \dfrac{{4 + 1 - 2}}{2}
⇒2tan⁡2θ+sin⁡2θ−1=32\Rightarrow 2{\tan ^2}\theta + {\sin ^2}\theta - 1 = \dfrac{3}{2}

∴\therefore The value of 2tan⁡2θ+sin⁡2θ−12{\tan ^2}\theta + {\sin ^2}\theta - 1 is 32\dfrac{3}{2}

Note: * The table that tells us some basic values of trigonometric functions at common angles is given as

Angles (in degrees)0∘{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}11
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
cosecNot defined22\sqrt 2 23\dfrac{2}{{\sqrt 3 }}1
sec123\dfrac{2}{{\sqrt 3 }}2\sqrt 2 2Not defined
cotNot defined3\sqrt 3 113\dfrac{1}{{\sqrt 3 }}0
  • Inverse of any function when applied on the same function cancels out the function and the inverse, i.e. f−1(f(x))=x{f^{ - 1}}(f(x)) = x.