Question

Write an expression using all six trigonometric functions such that the value of the expression is $3$.

Answer 1

Finding the required expression.

And we know these basic relations,
$$\begin{gathered} si{n}^2\left(x\right)+co{s}^2\left(x\right)=1 \\ tan\left(x\right)\times cot\left(x\right)=1 \\ cosec\left(x\right)\times sin\left(x\right)=1 \\ sec\left(x\right)\times cos\left(x\right)=1 \end{gathered}$$

So, we can write:
$$\begin{gathered} si{n}^2\left(x\right)+co{s}^2\left(x\right)+cosec\left(x\right)\times sin\left(x\right)+sec\left(x\right)\times cos\left(x\right)+tan\left(x\right)\times cot\left(x\right)-1 \\ =1+1+1+1-1 \\ =3 \end{gathered}$$

We can see that the above equation contains all the six trigonometric ratios.

Hence, the required equation is $si{n}^2\left(x\right)+co{s}^2\left(x\right)+cosec\left(x\right)\times sin\left(x\right)+sec\left(x\right)\times cos\left(x\right)+tan\left(x\right)\times cot\left(x\right)-1=3$