find the trigonometric ratios of 900 degree
Answers
Answer:
since 900 =1 and cos 90 degrees =0
Step-by-step explanation:
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¯¯¯¯¯¯¯¯
= 0OP¯¯¯¯¯¯¯¯, [since, θ → 0° therefore, OQ¯¯¯¯¯¯¯¯ → 0].
= 0
Therefore cos 90° = 0
limθ→90° tan θ
= limθ→90°PQ¯¯¯¯¯¯¯¯OQ¯¯¯¯¯¯¯¯¯
= OP¯¯¯¯¯¯¯¯0 [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.