Question

Prove that \[\dfrac{{\csc ({{90}^ \circ } + x) + \cot ({{450}^ \circ } + x)}}{{\csc ({{90}^ \circ } - x) + \tan ({{180}^ \circ } - x)}} + \dfrac{{\tan ({{180}^ \circ } + x) + \sec ({{180}^ \circ } - x)}}{{\tan ({{360}^ \circ } + x) - \sec ( - x)}} = 2\].

Answer 1

Hint: We will first simplify the LHS using the formulas of \[\csc ({90^ \circ } \pm x)\], \[\cot ({360^ \circ } + x)\], \[\tan ({180^ \circ } \pm x)\], \[\sec ({180^ \circ } - x)\], \[\sec ( - x)\] and \[\tan ({360^ \circ } + x)\].
Then, put in the values of known trigonometric functions and thus we will get our answer.

Complete step-by-step answer:
Before writing the required identities and formulas required for the solution first, we must know that in I quadrant, all the trigonometric functions possess positive value, in II quadrant, only sine and cosecant take positive values, in III quadrant, tangent and cotangent possess positive values, and in IV quadrant, cosine and secant takes positive values.
Let us now write the required formulas:
\[\csc ({90^ \circ } + x) = \sec (x)\] (It is positive because cosecant is positive in II quadrant )
\[\csc ({90^ \circ } - x) = \sec x\] (In I quadrant, all the trigonometric functions possess positive value)
\[\cot ({360^ \circ } + x) = \cot x\] (In I quadrant, all the trigonometric functions possess positive value)
\[\tan ({180^ \circ } + x) = \tan x\] (In III quadrant, tangent possess positive values)
\[\tan ({180^ \circ } - x) = - \tan x\] (There is a negative sign because tangent is negative in II quadrant)
\[\sec ({180^ \circ } - x) = - \sec x\] (There is a negative sign because secant is negative in II quadrant )
\[\sec ( - x) = \sec x\] (In IV quadrant, secant takes positive values)
\[\tan ({360^ \circ } + x) = \tan x\] (In I quadrant, tangent takes positive values)
$\cot ({90^ \circ } + x) = - \tan x$ (In II quadrant, cotangent is negative)
Putting all of these values in the LHS of the given expression:\[\dfrac{{\csc ({{90}^ \circ } + x) + \cot ({{450}^ \circ } + x)}}{{\csc ({{90}^ \circ } - x) + \tan ({{180}^ \circ } - x)}} + \dfrac{{\tan ({{180}^ \circ } + x) + \sec ({{180}^ \circ } - x)}}{{\tan ({{360}^ \circ } + x) - \sec ( - x)}}\] ……(2)
It becomes:
\[\dfrac{{\sec x + \cot ({{360}^ \circ } + ({{90}^ \circ } + x))}}{{\sec x - \tan x}} + \dfrac{{\tan x - \sec x}}{{\tan x - \sec x}}\] ……(2)
We see that \[\dfrac{{\tan x - \sec x}}{{\tan x - \sec x}} = 1\]
So, (2) becomes:
\[\dfrac{{\sec x + \cot ({{360}^ \circ } + ({{90}^ \circ } + x))}}{{\sec x - \tan x}} + 1\]
Now using formula number 3) in this, we get:
\[\dfrac{{\sec x + \cot ({{90}^ \circ } + x)}}{{\sec x - \tan x}} + 1\]
Using formula number 9) in this, we will get:-
\[\dfrac{{\sec x - \tan x}}{{\sec x - \tan x}} + 1\]
Simplifying it, we get:-
$1 + 1 = 2$, which is equal to RHS.
Hence, LHS = RHS.
Hence, we have proved the required expression.

Note: The student might make the mistake of a negative or positive sign, there are a lot of trigonometric functions involved in the question.
If there is a problem, remembering the signs of all the trigonometric functions, we must try to make a right angled triangle in the respective quadrant with known side lengths and just find the trigonometric value of the angle. See if it is negative or positive.