Advertisements
Advertisements
प्रश्न
Find the value of the following:
tan 15° tan 30° tan 45° tan 60° tan 75°
Advertisements
उत्तर १
tan 30° = `1/sqrt(3)`, tan 45° = 1, tan 60° = `sqrt(3)`
tan 15°. tan 30°. tan 45°. tan 60°. tan 75° = `tan 15^circ * 1/sqrt(3) * 1 * sqrt(3) tan 75^circ`
= `tan 15^circ xx tan 75^circ xx 1/sqrt(3) xx 1 xx sqrt(3)`
= `tan(90^circ - 75^circ) xx 1/(cot75^circ) xx 1` ...[tan 90° – θ = cot θ]
= `cot 75^circ xx 1/(cot75^circ) xx 1`
= 1
उत्तर २
Step-by-step values:
-
tan 15∘ = 2 − √3
-
tan 30∘ = `1/sqrt3`
-
tan 45∘ = 1
-
tan 60∘ = √3
-
tan 75∘ = 2 + √3
`tan 15° xx tan 75° = (2-sqrt3)(2+sqrt3)=4-3=1`
`tan 30°xxtan60° = 1/sqrt3 xx sqrt3 = 1`
tan 45∘ = 1
1 × 1 × 1
= 1
APPEARS IN
संबंधित प्रश्न
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
if `tan theta = 12/5` find the value of `(1 + sin theta)/(1 -sin theta)`
Evaluate:
3cos80° cosec10° + 2 sin59° sec31°
Evaluate:
3 cos 80° cosec 10° + 2 cos 59° cosec 31°
If 0° < A < 90°; find A, if `(cos A )/(1 - sin A) + (cos A)/(1 + sin A) = 4`
If θ is an acute angle such that \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\] \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\]
If 8 tan x = 15, then sin x − cos x is equal to
If 3 cos θ = 5 sin θ, then the value of
Express the following in term of angles between 0° and 45° :
cosec 68° + cot 72°
Find the value of the following:
sin 21° 21′
