Trigonometrical Ratios of 90°

How to Find the Trigonometrical Ratios of 90°?

Let a rotating line  \(\overrightarrow{OX}\) rotates about O in the anti-clockwise sense and starting from its initial position \(\overrightarrow{OX}\) traces out ∠XOY = θ where θ is very nearly equal to 90°.

Trigonometrical Ratios of 90°













Let \(\overrightarrow{OX}\) ⊥ \(\overrightarrow{OZ}\) therefore, ∠XOZ = 90°





Take a point P on \(\overrightarrow{OY}\) and draw \(\overline{PQ}\) perpendicular to \(\overline{OX}\).

Then,

Sin θ = \(\frac{\overline{PQ}}{\overline{OP}}\);

cos θ = \(\frac{\overline{OQ}}{\overline{OP}}\)

and tan θ =\(\frac{\overline{PQ}}{\overline{OQ}}\)

 

When θ is slowly approaches 90° and finally tends to 90° then,

(a) \(\overline{OQ}\) slowly decreases and finally tends to zero and

(b) the numerical difference between \(\overline{OP}\)  and \(\overline{PQ}\)  becomes very small and finally tends to zero.

Hence, in the Limit when θ → 90° then \(\overline{OQ}\) → 0 and \(\overline{PQ}\)   → \(\overline{OP}\)  . Therefore, we get

\(\lim_{θ \rightarrow 90°} \) sin  θ

= \(\lim_{θ \rightarrow 90°}\frac{\overline{PQ}}{\overline{OP}} \)

= \(\frac{\overline{OP}}{\overline{OP}} \) [since, θ → 90° therefore, \(\overline{PQ}\)   → \(\overline{OP}\) ].

= 1

Therefore sin 90° = 1

 

\(\lim_{θ \rightarrow 90°} \) cos θ

= \(\lim_{θ \rightarrow 90°}\frac{\overline{OQ}}{\overline{OP}} \)

= \(\frac{0}{\overline{OP}} \), [since, θ → 0° therefore, \(\overline{OQ}\) → 0].

= 0

Therefore cos 90° = 0

 

\(\lim_{θ \rightarrow 90°}\) tan θ

= \(\lim_{θ \rightarrow 90°}\frac{\overline{PQ}}{\overline{OQ}}\)

= \(\frac{\overline{OP}}{0}\) [since, θ → 0° \(\overline{OQ}\) → 0 and \(\overline{PQ}\)   → \(\overline{OP}\)].

= undefined

Therefore tan 900 = undefined

 

Thus,

csc 90° = \(\frac{1}{sin  90°} \)

= \(\frac{1}{1} \), [since, sin 90° = 1] 

= 1

 

sec 90° = \(\frac{1}{cos  90°} \)

= \(\frac{1}{0} \), [since, cos  90° = 0] 

= undefined


cot 0° = \(\frac{ cos  90°}{ sin  90°} \)

= \(\frac{0}{1} \), [since, sin 900 = 1 and cos 90° = 0] 

= 0


Trigonometrical Ratios of 90 degree are commonly called standard angles and the trigonometrical ratios of these angles are frequently used to solve particular angles.





● Trigonometric Functions






11 and 12 Grade Math

From Trigonometrical Ratios of 90° to HOME PAGE


New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.



Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.