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 \[2\] .