Question

Evaluate the following:
$4{{a}^{2}}{{\sin }^{2}}\left( \dfrac{3\pi }{4} \right)-3{{\left[ a\tan {{225}^{\circ }} \right]}^{2}}+{{\left[ 2a\cos {{315}^{\circ }} \right]}^{2}}=$
$\left( a \right)0$
$\left( b \right)a$
$\left( c \right)\sqrt{2a}$
$\left( d \right){{a}^{2}}$

Answer 1

We start by simplifying each of the trigonometric ratios. Firstly, we will simplify $\sin \left( \dfrac{3\pi }{4} \right)$ using $\sin \left( \pi -\theta \right)=\sin \theta .$ Then to simplify for ${{225}^{\circ }}$ we will use $\tan \left( {{270}^{\circ }}-\theta \right)=\cot \theta$ and then we will use $\cos \left( {{360}^{\circ }}-\theta \right)=\cos \theta$ to simplify $\cos {{315}^{\circ }}.$ Then we will put the value of these ratios in the original equation and then simplify to get the required solution.

Complete step-by-step answer:
We are asked to find the value of $4{{a}^{2}}{{\sin }^{2}}\left( \dfrac{3\pi }{4} \right)-3{{\left[ a\tan {{225}^{\circ }} \right]}^{2}}+{{\left[ 2a\cos {{315}^{\circ }} \right]}^{2}}.$ To solve this, we have to go through some simplification. We know that,
$\sin \left( \pi -\theta \right)=\sin \theta$
So, we can write $\sin \left( \dfrac{3\pi }{4} \right)=\sin \left( \pi -\dfrac{\pi }{4} \right)$
This can be simplified using $\sin \left( \pi -\theta \right)=\sin \theta .$ Using $\theta =\dfrac{\pi }{4}$ as $\sin \left( \dfrac{\pi }{4} \right)$ we get,
$\sin \left( \dfrac{3\pi }{4} \right)=\sin \left( \dfrac{\pi }{4} \right)$
We know that, $\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}.$
So,
$\sin \dfrac{3\pi }{4}=\dfrac{1}{\sqrt{2}}.......\left( i \right)$
Now, we have $\tan {{225}^{\circ }}$ which can be written as,
$\tan \left( {{225}^{\circ }} \right)=\tan \left( {{270}^{\circ }}-{{45}^{\circ }} \right)$
Now, we can use the above formula, we will get,
$\tan \left( {{270}^{\circ }}-{{45}^{\circ }} \right)=\cot \left( {{45}^{\circ }} \right)$
Now, as $\cot {{45}^{\circ }}=1,$ we get,
$\tan {{225}^{\circ }}=1......\left( ii \right)$
We also have that $\cos \left( {{360}^{\circ }}-\theta \right)$ is given as $\cos \theta .$ So as we have $\cos {{315}^{\circ }},$ this can be written as
$\cos \left( {{315}^{\circ }} \right)=\cos \left( {{360}^{\circ }}-{{45}^{\circ }} \right)$
So, using the above formula, $\cos \left( {{360}^{\circ }}-\theta \right)=\cos \theta$ on $\cos {{315}^{\circ }},$ we get,
$\cos \left( {{315}^{\circ }} \right)=\cos \left( {{360}^{\circ }}-{{45}^{\circ }} \right)$
$\Rightarrow \cos \left( {{315}^{\circ }} \right)=\cos \left( {{45}^{\circ }} \right)$
As we know that,
$\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$
So, we get,
$\cos {{315}^{\circ }}=\dfrac{1}{\sqrt{2}}.....\left( iii \right)$
Now, we use the value of (i), (ii) and (iii) on $4{{a}^{2}}{{\sin }^{2}}\left( \dfrac{3\pi }{4} \right)-3{{\left[ a\tan {{225}^{\circ }} \right]}^{2}}+{{\left[ 2a\cos {{315}^{\circ }} \right]}^{2}}.$ We will get,
$4\times {{a}^{2}}{{\sin }^{2}}\left( \dfrac{3\pi }{4} \right)-3{{\left[ a\tan {{225}^{\circ }} \right]}^{2}}+{{\left[ 2a\cos {{315}^{\circ }} \right]}^{2}}$
On simplifying, we will get,
$\Rightarrow 2{{a}^{2}}-3{{a}^{2}}+{{\left( \sqrt{2}a \right)}^{2}}$
$\Rightarrow 2{{a}^{2}}-3{{a}^{2}}+2{{a}^{2}}$
$\Rightarrow {{a}^{2}}$
So, we get the value of $4{{a}^{2}}{{\sin }^{2}}\left( \dfrac{3\pi }{4} \right)-3{{\left[ a\tan {{225}^{\circ }} \right]}^{2}}+{{\left[ 2a\cos {{315}^{\circ }} \right]}^{2}}$ as ${{a}^{2}}.$

So, the correct answer is “Option d”.

Note: Remember that any trigonometry ratio remains as original ratio when we add or subtract angle theta from multiple of 180 and any trigonometry ratio changes to its alternate if we add or subtract from ${{90}^{\circ }}$ or ${{270}^{\circ }}$ angle. Also, remember that 2 is made up of two $\left( \sqrt{2} \right).$ So, $2a\times \dfrac{1}{\sqrt{2}}$ can be written as $\sqrt{2}\times \sqrt{2}\times a\times \dfrac{1}{\sqrt{2}}.$ So, after simplification, we get $\sqrt{2}a.$