Question

If $t={{45}^{\circ }}$ then what is the value of $\sec \left( t \right)\sin \left( t \right)-\csc \left( t \right)\cos \left( t \right)$?
(a) -2
(b) -1
(c) 0
(d) $\dfrac{\pi }{2}$
(e) None of the above

Answer 1

Assume the given expression as E and use the conversions $\sec x=\dfrac{1}{\cos x}$ and $\csc x=\dfrac{1}{\sin x}$ to simplify the expression. Now, convert the expression into the trigonometric expression containing the tangent and the co – tangent function by using the conversions $\dfrac{\sin x}{\cos x}=\tan x$ and $\dfrac{\cos x}{\sin x}=\cot x$. Finally, substitute the value of angle $t={{45}^{\circ }}$ and use the values $\tan {{45}^{\circ }}=1$ and $\cot {{45}^{\circ }}=1$ to get the answer.

Complete step-by-step solution:
Here we have been provided with the expression $\sec \left( t \right)\sin \left( t \right)-\csc \left( t \right)\cos \left( t \right)$ and we have to find its value for the provided angle $t={{45}^{\circ }}$. Let us assume the given expression as E, so we have,
$\Rightarrow E=\sec \left( t \right)\sin \left( t \right)-\csc \left( t \right)\cos \left( t \right)$
Using the conversions $\sec x=\dfrac{1}{\cos x}$ and $\csc x=\dfrac{1}{\sin x}$ we get,
$\begin{aligned} & \Rightarrow E=\dfrac{1}{\cos \left( t \right)}\times \sin \left( t \right)-\dfrac{1}{\sin \left( t \right)}\times \cos \left( t \right) \\\ & \Rightarrow E=\dfrac{\sin \left( t \right)}{\cos \left( t \right)}-\dfrac{\cos \left( t \right)}{\sin \left( t \right)} \\\ \end{aligned}$
We know that $\dfrac{\sin x}{\cos x}=\tan x$ and $\dfrac{\cos x}{\sin x}=\cot x$, so we can write the above expression as: -
$\Rightarrow E=\tan \left( t \right)-\cot \left( t \right)$
Substituting the value of the given angle $t={{45}^{\circ }}$ and using the values $\tan {{45}^{\circ }}=1$ and $\cot {{45}^{\circ }}=1$ we get,

& \Rightarrow E=\tan \left( {{45}^{\circ }} \right)-\cot \left( {{45}^{\circ }} \right) \\\ & \Rightarrow E=1-1 \\\ & \therefore E=0 \\\ \end{aligned}$$ **Hence, option (c) is the correct answer.** **Note:** Note that you can also simplify the expression in a different way. You can take the L.C.M and write the expression as $\dfrac{{{\sin }^{2}}\left( t \right)-{{\cos }^{2}}\left( t \right)}{\sin \left( t \right)\cos \left( t \right)}$ and then use the identity ${{\cos }^{2}}\left( t \right)-{{\sin }^{2}}\left( t \right)=\cos \left( 2t \right)$ to simplify the numerator. Then we will put the value of $t={{45}^{\circ }}$ and use the value $\cos {{90}^{\circ }}=0$ to get the answer. You must remember the values of all the trigonometric function for some particular angles like ${{0}^{\circ }},{{30}^{\circ }},{{45}^{\circ }},{{60}^{\circ }}$ and ${{90}^{\circ }}$. In higher trigonometry we also have to learn the trigonometric values of angles ${{18}^{\circ }},{{36}^{\circ }},{{54}^{\circ }}$ and ${{72}^{\circ }}$. Also remember an important identity $2\sin \left( t \right)\cos \left( t \right)=\sin \left( 2t \right)$ in case you want to simplify the denominator and use the value $\sin {{90}^{\circ }}=1$.