Question

Which of the following is possible?
1. $\sin \theta = \dfrac{{({a^2} + {b^2})}}{{({a^2} - {b^2})}}$ , $(a \ne b)$
2. $\sec \theta = \dfrac{4}{5}$
3. $\tan \theta = 45$
4. $\cos \theta = \dfrac{7}{3}$

Answer 1

Hint : In this question, we have to find which of the given options is possible or true. We will check one by one each option to find the correct option.
Given options are based on values of Trigonometric ratios. We will use the fact that values of sine and cosine are always less than $1$ and the value of tangent is always between $0$ to infinity. Secant is reciprocal of cosine.

Complete step-by-step answer :
Given question is based on trigonometric ratio. Trigonometry is the branch of mathematics which deals with relationships between side length and angles in any right angle triangle. Trigonometric ratio is the ratio of two sides of a right angle triangle.
Let ABC be a right angle triangle such that angle $\angle b = 90$ . Then the side opposite to the right angle is called the hypotenuse. While the other two are perpendicular and base depending on which angle we are finding the T-ratio.
Considering the given question, we will check each option one by one.
We are given, $\sin \theta = \dfrac{{({a^2} + {b^2})}}{{({a^2} - {b^2})}}$ , $(a \ne b)$
We know that the sum of squares of two numbers is always greater than the difference of squares of the same two numbers. i.e. $({a^2} + {b^2}) > ({a^2} - {b^2})$
On dividing both side by$({a^2} - {b^2})$ we get,
$\dfrac{{({a^2} + {b^2})}}{{({a^2} - {b^2})}} > 1$
Hence we have $\sin \theta > 1$
But the value of sine always lies between $- 1$ and $1$.
Hence option (1) is not possible.
Now given, $\sec \theta = \dfrac{4}{5}$
Taking reciprocal we have $\dfrac{1}{{\sec \theta }} = \dfrac{5}{4}$
We know that $\cos \theta = \dfrac{1}{{\sec \theta }}$
Hence we have $\cos \theta = \dfrac{5}{4} > 1$
But the value of cosine always lies between $- 1$ and $1$ .
Hence option (2) is not possible.
Now given, $\tan \theta = 45$
We know that the value of the tangent lies between $0$ and infinity.
Hence option (3) is possible.
Now given $\cos \theta = \dfrac{7}{3}$
Since $\dfrac{7}{3} > 1$
But value of cosine always lie between $- 1$ and $1$
Hence option (4) is also not possible.
Hence option (3) is the correct answer.
So, the correct answer is “Option 3”.

Note : T-Ratio is the ratio of two sides of a triangle.
The sum of squares of two numbers is always greater than the difference of squares of the same two numbers. i.e. $({a^2} + {b^2}) > ({a^2} - {b^2})$ .
Value of sine and cosine always lies between $- 1$ and $1$ .
Value of the tangent lies between $0$ and infinity.