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

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