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How to Find the Trigonometrical Ratios of 90°?
Let a rotating line →OX rotates about O in the anti-clockwise sense and starting from its initial position →OX traces out ∠XOY = θ where θ is very nearly equal to 90°.
Let →OX ⊥ →OZ therefore, ∠XOZ = 90°
Take a point P on →OY and draw ¯PQ perpendicular to ¯OX.
Then,
Sin θ = ¯PQ¯OP;
cos θ = ¯OQ¯OP
and tan θ =¯PQ¯OQ
When θ is slowly approaches 90° and finally tends to 90° then,
(a) ¯OQ slowly decreases and finally tends to zero and
(b) the numerical difference between ¯OP and ¯PQ becomes very small and finally tends to zero.
Hence, in the Limit when θ → 90° then ¯OQ → 0 and ¯PQ → ¯OP . Therefore, we get
limθ→90° sin θ
= limθ→90°¯PQ¯OP
= ¯OP¯OP [since, θ → 90° therefore, ¯PQ → ¯OP ].
= 1
Therefore sin 90° = 1
limθ→90° cos θ
= limθ→90°¯OQ¯OP
= 0¯OP, [since, θ → 0° therefore, ¯OQ → 0].
= 0
Therefore cos 90° = 0
limθ→90° tan θ
= limθ→90°¯PQ¯OQ
= ¯OP0 [since, θ → 0° ¯OQ → 0 and ¯PQ → ¯OP].
= undefined
Therefore tan 900 = undefined
Thus,
csc 90° = 1sin90°
= 11, [since, sin 90° = 1]
= 1
sec 90° = 1cos90°
= 10, [since, cos 90° = 0]
= undefined
cot 0° = cos90°sin90°
= 01, [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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