How to Find the Trigonometrical Ratios of 90°?
Let a rotating line \(\overrightarrow{OX}\) rotates about O in the anticlockwise sense and starting from its initial position \(\overrightarrow{OX}\) traces out ∠XOY = θ where θ is very nearly equal to 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.
11 and 12 Grade Math
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