Advertisements
Advertisements
प्रश्न
If A and B are acute angles such that tan A =`1/3, tan B = 1/2 and tan (A + B) =` show that `A+B = 45^0`
उत्तर
Given:
tan A = `1/3 and tan B = 1/2`
tan (A+B) = `(tan A + tan B)/(1- tan A tan B)`
On substituting these values in RHS of the expression, we get:
`(tan A + tan B )/(1- tan A tan B) = ((1/3 +1/2))/((1-1/3xx1/3)` =`((5/6))/(1-1/6) = ((5/6))/((5/6))=1`
⇒ tan (A + B) = 1= tan `45^0` [ ∵ tan 450 =1]
∴ A+B = 450
APPEARS IN
संबंधित प्रश्न
If 8 tan A = 15, find sin A – cos A.
If `sin (A – B) = 1/2` and `cos (A + B) = 1/2`, `0^@` < A + `B <= 90^@`, A > B Find A and B.
If tan θ = `1/sqrt(7) `show that ` (cosec ^2 θ - sec^2 θ)/(cosec^2 θ + sec^2 θ ) = 3/4`
If ∠A and ∠B are acute angles such that sin A = Sin B prove that ∠A = ∠B.
Evaluate:
cos450 cos300 + sin450 sin300
In the adjoining figure, ΔABC is a right-angled triangle in which ∠B = 900, ∠300 and AC = 20cm. Find (i) BC, (ii) AB.
In the adjoining figure, ΔABC is right-angled at B and ∠A = 450. If AC = 3`sqrt(2)`cm, find (i) BC, (ii) AB.
tan 30° × tan ______° = 1
Given: 4 cot A = 3
find :
(i) sin A
(ii) sec A
(iii) cosec2A - cot2A.
In ΔABC, ∠B = 90°. If AB = 12units and BC = 5units, find: tan A
