Question

If ${\text{cosec}}A = \dfrac{{15}}{7}$ and $A + B = {90^ \circ }$, find the value of $\sec B$.

Answer 1

Hint: First of all, find the value of $A$ in terms of $B$ from the given relation, that, $A + B = {90^ \circ }$. Next, substitute the value of $A$ in the formula, ${\text{cosec}}A = \dfrac{{15}}{7}$ and then use the identity that ${\text{cosec}}\left( {{{90}^ \circ } - \theta } \right) = \sec \theta $ to write the value of $\sec B$.

Complete step-by-step answer:
We are given that ${\text{cosec}}A = \dfrac{{15}}{7}$ and $A + B = {90^ \circ }$.
We have to find the value of $\sec B$.
First of all, let us find the value of $A$ in terms of $B$
That is, $A = {90^ \circ } - B$
Let us now substitute the value of $A$ in the given value ${\text{cosec}}A = \dfrac{{15}}{7}$
Hence, we have ${\text{cosec}}\left( {{{90}^ \circ } - B} \right) = \dfrac{{15}}{7}$
As it is known that cosecant and secant are complementary ratios.
That is, ${\text{cosec}}\left( {{{90}^ \circ } - \theta } \right) = \sec \theta $ and ${\text{sec}}\left( {{{90}^ \circ } - \theta } \right) = \operatorname{cosec} \theta $
Therefore, we can write ${\text{cosec}}\left( {{{90}^ \circ } - B} \right) = \sec B$
And hence the value of $\sec B = \dfrac{{15}}{7}$

Note: The complementary ratios in trigonometry are, sine and cosine, tangent and cotangent, and cosecant and secant. That is, $\sin \left( {90 - x} \right) = \cos x$, $\tan \left( {90 - x} \right) = \cot x$ and so on. Here, the value of ${\text{cosec}}A$ and $\sec B$ are the same because $A + B = {90^ \circ }$.