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,
   $cosec\theta = \dfrac{1}{\sin \theta}$ and $\sec \theta = \dfrac{1}{{\cos \theta }}$
   So , $cosec\theta - \sec \theta = cosec\theta \times \sec \theta$ can be written as
   $\dfrac{1}{{sin\theta }} - {\text{ }}\dfrac{1}{{cos{\text{ }}\theta }}{\text{ }} = {\text{ }}\dfrac{1}{{\left( {{\text{ }}sin{\text{ }}\theta {\text{ }} \times {\text{ }}cos{\text{ }}\theta {\text{ }}} \right)}}$
   On , simplifying we get
   $\dfrac{{\cos \theta - \sin \theta }}{{\sin \theta \cos \theta }} = \dfrac{1}{{\sin \theta \cos \theta }}$
   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 ,
   $\dfrac{1}{{\left( {{\text{ }}sin{\text{ }}\theta {\text{ }} \times {\text{ }}cos{\text{ }}\theta {\text{ }}} \right)}}{\text{ }} = {\text{ }}1$
   $1 = \sin \theta \times \cos \theta$
   Multiplying both side by 2 , we get
   $2 = 2\sin \theta \times \cos \theta$
   We know that ${{\text{ }}sin{\text{ }}2x{\text{ }} = {\text{ }}2{\text{ }}sin{\text{ }}x{\text{ }} \times {\text{ }}cos{\text{ }}x{\text{ }}}$
   $2 = \sin 2\theta$
   As , we know that the value of sin functions lies in the interval $\left[ { - 1{\text{ }},{\text{ }}1{\text{ }}} \right]$
   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 $sin 2\theta {\text{ }} = {\text{ }}2$has no solution .

$3)$ $\cos \theta + \sin \theta = \sqrt 2$
Dividing the equation$3)$ by $\sqrt 2$ , we get
$\dfrac{{\cos \theta }}{{\sqrt 2 }} + \dfrac{{\sin \theta }}{{\sqrt 2 }} = 1$
We also know that the value of the trigonometric function $\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ and $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$
So , we get the equation as
$\sin \theta \times \cos \dfrac{\pi }{4} + \cos \theta \times \sin \dfrac{\pi }{4} = 1$
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
$\sin \left( {\theta + \dfrac{\pi }{4}} \right) = \sin \theta \times \cos \dfrac{\pi }{4} + \sin \dfrac{\pi }{4} \times \cos \theta$
So , we get
$\sin \left( {\theta + \dfrac{\pi }{4}} \right) = 1$
At $\theta = \dfrac{\pi }{4}$ , it has a solution

$4)$ $\sqrt 3 \sin \theta - \cos \theta = 2$
Dividing both sides by $2$ , the equation becomes
$\dfrac{{\sqrt 3 }}{2}\sin \theta - \dfrac{1}{2}\cos \theta = 1$
We also know that the value of the trigonometric function $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$ and $\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}$
So , we get the equation as
$\sin \theta \times \cos \dfrac{\pi }{6} - \cos \theta \times \sin \dfrac{\pi }{6} = 1$
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
$\sin \left( {\theta - \dfrac{\pi }{6}} \right) = \sin \theta \times \cos \dfrac{\pi }{6} - \sin \dfrac{\pi }{6} \times \cos \theta$
So , we get
$\sin \left( {\theta - \dfrac{\pi }{6}} \right) = 1$
At $\theta = \dfrac{{2\pi }}{3}$ , 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 $2$ .