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
if `cot theta = 1/sqrt3` find the value of `(1 - cos^2 theta)/(2 - sin^2 theta)`
solve.
sec2 18° - cot2 72°
Evaluate.
`cot54^@/(tan36^@)+tan20^@/(cot70^@)-2`
Evaluate.
`(2tan53^@)/(cot37^@)-cot80^@/tan10^@`
Evaluate.
`cos^2 26^@+cos65^@sin26^@+tan36^@/cot54^@`
For triangle ABC, show that : `sin (A + B)/2 = cos C/2`
Prove that:
`1/(1 + cos(90^@ - A)) + 1/(1 - cos(90^@ - A)) = 2cosec^2(90^@ - A)`
If A and B are complementary angles, prove that:
cot A cot B – sin A cos B – cos A sin B = 0
If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ, then 2 sin 3θ −\[\sqrt{3} \tan 3\theta\] is equal to
If x tan 45° sin 30° = cos 30° tan 30°, then x is equal to ______.
