Question

Prove that:
$\left( a \right)\dfrac{cos\left( \pi +\theta \right)cos\left( -\theta \right)}{sin\left( \pi -\theta \right)cos\left( \dfrac{\pi }{2}+\theta \right)}=co{{t}^{2}}\theta$
$\left( b \right)\cos \left( \theta \right)+\text{sin}\ \ \left( {{270}^{{}^\circ }}+\theta \right)-\text{sin}\ \ \left( {{270}^{{}^\circ }}-\theta \right)+\text{cos}\ \ \left( {{180}^{{}^\circ }}+\theta \right)=0$
$\left( c \right)cos\left( \dfrac{3\pi }{2}+\theta \right)cos\left( 2\pi +\theta \right)\left[ cot\left( \dfrac{3\pi }{2}-\theta \right)+cot\left( 2\pi +\theta \right) \right]=1$

Answer 1

We are given two expressions involving trigonometric terms in each part and we need to prove that one is equal to the other. For this we use several trigonometric formulae and we try to simplify the equation as much as possible. We will start with the left side and we then perform some operations on it to turn it into the expression written on the right hand side.

Complete step-by-step solution:
Consider
$\left( a \right)\dfrac{cos\left( \pi +\theta \right)cos\left( -\theta \right)}{sin\left( \pi -\theta \right)cos\left( \dfrac{\pi }{2}+\theta \right)}=co{{t}^{2}}\theta$
We start with the left hand side, we have
$LHS=\dfrac{cos\left( \pi +\theta \right)cos\left( -\theta \right)}{sin\left( \pi -\theta \right)cos\left( \dfrac{\pi }{2}+\theta \right)}$
We see that inside the angle of cosine we have $\pi+\theta$ and $\dfrac{\pi}{2}+\theta$ type terms, so we use the following formulae:
$cos(\pi +\theta )=-\cos (\theta )$
$cos(-\theta )=\cos (\theta )$
$\sin (\pi -\theta )=\sin (\theta )$
$cos\left( \dfrac{\pi }{2}+\theta \right)=-\sin (\theta )$
Putting these in LHS, we get
$\dfrac{-\cos \left( \theta \right)cos\left( \theta \right)}{sin\left( \theta \right)\left( -\sin \left( \theta \right) \right)}=\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }={{\left( \dfrac{\cos \theta }{\sin \theta } \right)}^{2}}$
Now, we use the following formula:
$cot(A)=\dfrac{cos(A)}{sin(A)}$
We get:
${{\left( \dfrac{\cos \theta }{\sin \theta } \right)}^{2}}={{\cot }^{2}}\theta =RHS$
Hence, proved.
Consider
$\left( b \right)\cos \left( \theta \right)+\text{sin}\ \ \left( {{270}^{{}^\circ }}+\theta \right)-\text{sin}\ \ \left( {{270}^{{}^\circ }}-\theta \right)+\text{cos}\ \ \left( {{180}^{{}^\circ }}+\theta \right)=0$
We start with the left hand side, we have
$LHS=\cos \left( \theta \right)+\text{sin}\ \ \left( {{270}^{{}^\circ }}+\theta \right)-\text{sin}\ \ \left( {{270}^{{}^\circ }}-\theta \right)+\text{cos}\ \ \left( {{180}^{{}^\circ }}+\theta \right)$
We know that
${{270}^{{}^\circ }}=\dfrac{3\pi }{2}$
${{180}^{{}^\circ }}=\pi$
So, we get:
$\cos \left( \theta \right)+\text{sin}\ \ \left( \dfrac{3\pi }{2}+\theta \right)-\text{sin}\ \ \left( \dfrac{3\pi }{2}-\theta \right)+\text{cos}\ \ \left( \pi +\theta \right)$
We use the formulae below:
$\text{sin}\ \ \left( \dfrac{3\pi }{2}+\theta \right)=-\cos \left( \theta \right)$
$\text{sin}\ \ \left( \dfrac{3\pi }{2}-\theta \right)=-\cos \left( \theta \right)$
$\text{cos}\ \ \left( \pi +\theta \right)=-\cos \left( \theta \right)$
Putting the values in the expression, we get:
$\cos \left( \theta \right)-\cos \left( \theta \right)-\left( -\cos \left( \theta \right) \right)-\cos \left( \theta \right)=\cos \left( \theta \right)-\cos \left( \theta \right)+\cos \left( \theta \right)-\cos \left( \theta \right)=0=RHS$
Hence, proved.
Consider
$\left( c \right)cos\left( \dfrac{3\pi }{2}+\theta \right)cos\left( 2\pi +\theta \right)\left[ cot\left( \dfrac{3\pi }{2}-\theta \right)+cot\left( 2\pi +\theta \right) \right]=1$
Again, we use the formulae used in the previous questions. We will use these:
$\cos \left( \dfrac{3\pi }{2}+\theta \right)=\sin \left( \theta \right)$
$\text{cos}\left( 2\pi +\theta \right)=\cos \left( \theta \right)$
$\cot \left( \dfrac{3\pi }{2}-\theta \right)=\tan \left( \theta \right)$
$\text{cot}\left( 2\pi +\theta \right)=\cot \left( \theta \right)$
After plugging these we get:
$\sin \left( \theta \right)\cos \left( \theta \right)\left[ \tan \left( \theta \right)+\cot \left( \theta \right) \right]$
Now, we use the following formulae:
$tan(A)=\dfrac{sin(A)}{cos(A)}$
$cot(A)=\dfrac{cos(A)}{sin(A)}$
After putting these we get:
$\sin \left( \theta \right)\cos \left( \theta \right)\left[ \dfrac{sin(\theta )}{cos(\theta )}+\dfrac{cos(\theta )}{sin(\theta )} \right]$
$\Rightarrow \sin \left( \theta \right)\cos \left( \theta \right)\left[ \dfrac{si{{n}^{2}}(\theta )+co{{s}^{2}}(\theta )}{cos(\theta )sin(\theta )} \right]$
Now, we use
$si{{n}^{2}}(A)+co{{s}^{2}}(A)=1$
And this holds as true for any$A$. So, we get:
$\sin \left( \theta \right)\cos \left( \theta \right)\left[ \dfrac{1}{cos(\theta )sin(\theta )} \right]=1=RHS$
Hence, proved.

Note: Most of the time, while using the trigonometric formulae, we may plug in the wrong equations. So, the trigonometric formulae must be learned properly so that you are able to prove the quality of expressions without any mistakes. Moreover, always start with the side that is complex. As in this question, if you start with the left hand side then you will have no operation to perform and you will get stuck, so always start with the complex side and try to reduce it to the expression that doesn’t involve much terms.