Question

Which one of the following equations has no solution?
A. $cosec\theta -\sec \theta =cosec\theta \times \sec \theta$
B. $cosec\theta \times \sec \theta =1$
C. $\cos \theta +\sin \theta =\sqrt{2}$
D. $\sqrt{3}\sin \theta -\cos \theta =2$

Answer 1

Hint: We have to find which of the following equations has no solution . We solve this question using the concept of general trigonometric solutions . Solving the equations in the options we can find which of the equations has a solution or not . For that we can take each option and solve it and check it has solutions in its defined intervals.

Complete step-by-step answer:
All the trigonometric functions are classified into two categories or types as either sine function or cosine function . All the functions which lie in the category of sine functions are sin , cosec and tan functions on the other hand the functions which lie in the category of cosine functions are cos , sec and cot functions . The trigonometric functions are classified into these two categories on the basis of their property which is stated as : when the value of angle is substituted by the negative value of the angle then we get the negative value for the functions in the sine family and a positive value for the functions in the cosine family .
Taking the first equation
1) $cosec\theta -\sec \theta =cosec\theta \times \sec \theta$
We know that,
 $cosec\theta ={\displaystyle \frac{1}{\sin \theta }}$ and $\sec \theta ={\displaystyle \frac{1}{\cos \theta }}$
So , $cosec\theta -\sec \theta =cosec\theta \times \sec \theta$ can be written as
$${\displaystyle \frac{1}{sin\theta }}-{\displaystyle \frac{1}{cos\theta }}={\displaystyle \frac{1}{\left(sin\theta \times cos\theta \right)}}$$
On , simplifying we get
 ${\displaystyle \frac{\cos \theta -\sin \theta }{\sin \theta \cos \theta }}={\displaystyle \frac{1}{\sin \theta \cos \theta }}$
Eliminating the denominator on both sides,
 $\cos \theta -\sin \theta =1$
 $\cos \theta =1+\sin \theta$
At $\theta =0$ , it has a solution.

2) $cosec\theta \times \sec \theta =1$
As done in the above case, we need to substitute the values as above.
So ,
$${\displaystyle \frac{1}{\left(sin\theta \times cos\theta \right)}}=1$$
 $1=\sin \theta \times \cos \theta$
Multiplying both side by 2 , we get
 $2=2\sin \theta \times \cos \theta$
We know that $$sin2x=2sinx\times cosx$$
 $2=\sin 2\theta$
As , we know that the value of sin functions lies in the interval $$\left[-1,1\right]$$
So , the maximum value of sin function is 1 .
Therefore for no value of $\theta$ , $\sin \theta$ can be equal to 2 .
Thus , the equation $$sin2\theta =2$$has no solution .

$$3)$$ $\cos \theta +\sin \theta =\sqrt{2}$
Dividing the equation$$3)$$ by $\sqrt{2}$ , we get
 ${\displaystyle \frac{\cos \theta }{\sqrt{2}}}+{\displaystyle \frac{\sin \theta }{\sqrt{2}}}=1$
We also know that the value of the trigonometric function $$\sin {\displaystyle \frac{\pi }{4}}={\displaystyle \frac{1}{\sqrt{2}}}$$ and $$\cos {\displaystyle \frac{\pi }{4}}={\displaystyle \frac{1}{\sqrt{2}}}$$
So , we get the equation as
$$\sin \theta \times \cos {\displaystyle \frac{\pi }{4}}+\cos \theta \times \sin {\displaystyle \frac{\pi }{4}}=1$$
We also know that the formula of sine function for sum of two terms i.e. $$\sin \left(a+b\right)=\sin a\times \cos b+\sin b\times \cos a$$
Using this formula , we get the above equation as
$$\sin \left(\theta +{\displaystyle \frac{\pi }{4}}\right)=\sin \theta \times \cos {\displaystyle \frac{\pi }{4}}+\sin {\displaystyle \frac{\pi }{4}}\times \cos \theta$$
So , we get
$$\sin \left(\theta +{\displaystyle \frac{\pi }{4}}\right)=1$$
At $\theta ={\displaystyle \frac{\pi }{4}}$ , it has a solution

$$4)$$ $\sqrt{3}\sin \theta -\cos \theta =2$
Dividing both sides by $2$ , the equation becomes
 ${\displaystyle \frac{\sqrt{3}}{2}}\sin \theta -{\displaystyle \frac{1}{2}}\cos \theta =1$
We also know that the value of the trigonometric function $$\sin {\displaystyle \frac{\pi }{6}}={\displaystyle \frac{1}{2}}$$ and $$\cos {\displaystyle \frac{\pi }{6}}={\displaystyle \frac{\sqrt{3}}{2}}$$
So , we get the equation as
$$\sin \theta \times \cos {\displaystyle \frac{\pi }{6}}-\cos \theta \times \sin {\displaystyle \frac{\pi }{6}}=1$$
We also know that the formula of sine function for difference of two terms i.e. $$\sin \left(a-b\right)=\sin a\times \cos b-\sin b\times \cos a$$
Using this formula , we get the above equation as
$$\sin \left(\theta -{\displaystyle \frac{\pi }{6}}\right)=\sin \theta \times \cos {\displaystyle \frac{\pi }{6}}-\sin {\displaystyle \frac{\pi }{6}}\times \cos \theta$$
So , we get
$$\sin \left(\theta -{\displaystyle \frac{\pi }{6}}\right)=1$$
At $\theta ={\displaystyle \frac{2\pi }{3}}$ , it has a solution
Hence , the correct option is (B) .
So, the correct answer is “Option B”.

Note: Here the Option $$(3)$$ has a solution at an angle $${45}^{\circ }$$ which can be calculated by dividing the LHS by $$\sqrt{2}$$ and the option $$(4)$$ has a solution at an angle $${120}^{\circ }$$ which can be calculated by dividing the LHS by $$2$$ .