Question

If $\sin \theta + \cos \theta = m$ and $\sec \theta + \cos ec\theta = n$, then $n(m + 1)(m - 1) =$
A.$m$
B.$n$
C.$2m$
D.$2n$

Answer 1

Hint : Trigonometric functions are also known as Circular Functions can be simply defined as the functions of an angle of a triangle. It means that the relationship between the angles and sides of a triangle are given by these trigonometric functions.

Complete step-by-step answer :
The word ‘trigonometry’ is derived from the Greek words ‘trigon’ and ‘metron’ and it means ‘measuring the sides of a triangle’. Trigonometric functions are also known as Circular Functions can be simply defined as the functions of an angle of a triangle. It means that the relationship between the angles and sides of a triangle are given by these trigonometric functions. The basic trigonometric functions are the sine, the cosine, the tangent, the cotangent, the secant and the cosecant.
The angles of the sine, the cosine, and the tangent are the primary classification of functions of trigonometry. And the three functions which are the cotangent, the secant and the cosecant can be derived from the primary functions.
Inverse trigonometric functions are used to obtain an angle from any of the angle’s trigonometric ratios. Basically, inverses of the sine, cosine, tangent, cotangent, secant, and cosecant functions are represented as arcsine, arccosine, arctangent, arc cotangent, arc secant, and arc cosecant.
We are given $\sin \theta + \cos \theta = m$
$\sec \theta + \cos ec\theta = n$
We are to find the value of $n(m + 1)(m - 1) =$ $n({m^2} - 1)$
Putting the respective given values we get ,
$= \left( {\sec \theta + \cos ec\theta } \right)\left[ {{{\left( {\sin \theta + \cos \theta } \right)}^2} - 1} \right]$
Opening the square term we get ,
$= \left( {\sec \theta + \cos ec\theta } \right)\left[ {{{\sin }^2}\theta + {{\cos }^2}\theta + 2\sin \theta \cos \theta - 1} \right]$
We know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$.
On simplification we get ,
$= \left( {\sec \theta + \cos ec\theta } \right)\left[ {2\sin \theta \cos \theta } \right]$
On further simplification we get ,
$= \sec \theta \left( {2\sin \theta \cos \theta } \right) + \cos ec\theta \left( {2\sin \theta \cos \theta } \right)$
$= 2\sin \theta + 2\cos \theta$
Taking $2$common we get ,
$= 2(\sin \theta + \cos \theta )$
$= 2m$
Therefore option ($3$) is the correct answer.
So, the correct answer is “Option 3”.

Note : The angles of the sine, the cosine, and the tangent are the primary classification of functions of trigonometry. And the three functions which are the cotangent, the secant and the cosecant can be derived from the primary functions.