Question

Express the ratios \[ \cos A \] , \[ \tan A \] and \[ \sec A \] in terms of \[ \sin A \] .

Answer 1

Hint: We need to express cosine, tangent and secant function only in terms of sine only. We use fundamental trigonometric identities. We also use the identity \[{ \sin ^2}A + { \cos ^2}A = 1 \] to solve these types of questions. Keep in mind that the solution that you get will be in terms of sine only and not any other function.

Complete step-by-step answer:
Now,
We express cosine in terms of sine.
We know the Pythagorean identity \[ \Rightarrow { \sin ^2}A + { \cos ^2}A = 1 \] .
Rearranging the terms we get,
  \[ \Rightarrow { \cos ^2}A = 1 - { \sin ^2}A \]
We need cosine term only, not its square, so simplifying we get,
  \[ \Rightarrow \cos A = \pm \sqrt {1 - {{ \sin }^2}A} \]
Here, \[A \] is acute. We know that cosine function is positive when \[A \] is acute.
So we take,
  \[ \Rightarrow \cos A = \sqrt {1 - {{ \sin }^2}A} \] . Which is in sine function only.
We express tangent in terms of sine
We know from fundamental trigonometric identity \[ \tan A = \dfrac{{ \sin A}}{{ \cos A}} \]
We can see that in the denominator we have cosine function. We convert it into a sine function.
From above we have,
  \[ \cos A = \sqrt {1 - {{ \sin }^2}A} \] . Substituting we get,
  \[ \Rightarrow \tan A = \dfrac{{ \sin A}}{{ \sqrt {1 - {{ \sin }^2}A} }} \] . Which is in sine function only.
We express secant in terms of sine.
We know from fundamental trigonometric identity \[ \sec A = \dfrac{1}{{ \cos A}} \] .
We can see that in the denominator we have cosine function. We convert it into a sine function. From above we have,
  \[ \cos A = \sqrt {1 - {{ \sin }^2}A} \] . Substituting we get,
  \[ \Rightarrow \sec A = \dfrac{1}{{ \sqrt {1 - {{ \sin }^2}A} }} \] . Which is in sine function only.

Note: Students need to remember all the fundamental trigonometric identity and also the Pythagorean identity. Knowing this we can solve these types of questions easily. In all the above solutions you can see that the final answer is in sine function only. They can also express them in cosine form also. Use the same procedure that was mentioned above.