Question

If tan(α - β) = $\frac{7}{24}$ and tan α = $\frac{4}{3}$, where α and β are in the first quadrant. Prove that α + β = $\frac{\pi}{2}$.

Answer 1

given, If tan(α - β) = $\frac{7}{24}$ and tan α = $\frac{4}{3}$, where α and β are in the first quadrant.

$tan(\alpha-\beta)=\frac{7}{24}$

and $tan\alpha=\frac{4}{3}$

we know, tan(A - B) = (tanA - tanB)/(1 + tanA. tanB)

so, $tan(\alpha-\beta)=\frac{7}{24}$

$\frac{tan\alpha-tan\beta}{1+tan\alpha tan\beta}=\frac{7}{24}$

$\frac{\frac{4}{3}-tan\beta}{1+\frac{4}{3}tan\beta}=\frac{7}{24}$

$\frac{4-3tan\beta}{3+4tan\beta}=\frac{7}{24}$

$24(4-3tan\beta)=7(3+4tan\beta)$

$96-72tan\beta=21+28tan\beta$

$96-21=72tan\beta+28tan\beta$

$75=100tan\beta$

$tan\beta=\frac{3}{4}$


now, $tan(\alpha+\beta)=\frac{tan\alpha+tan\beta}{1-tan\alpha tan\beta}$

= (3/4 + 4/3)/(1 - 3/4 × 4/3)

= (3/4 + 4/3)/0 = ∞ = tan90°

$\alpha+\beta=\frac{\pi}{2}$

Answer 2

HELLO DEAR,




GIVEN:-
tan(α - β) = 7/24
tanα = 4/3


we know, tan(A - B) = (tanA - tanB)/(1 + tanA. tanB)

so, (tanα - tanβ)/(1 + tanα. tanβ) = 7/24

=> (4/3 - tanβ)/(1 + (4/3). tanβ) = 7/24

=> 4 - 3tanβ = 3(7/24) + 4(7/24).tanβ

=> 4 - 7/8 = tanβ(18 + 7)/6

=> (32 - 7)/8 = tanβ.(25)/6

=> 25/8 = 25/6

=> 6/8 = tanβ

=> tanβ = 3/4


so, tan(α + β) = (tanα + tanβ)/(1 - tanα.tanβ)

=> (4/3 + 3/4)/(1 - 4/3.3/4)

=> (4/3 + 3/4)/0 = ∞ = tanπ/2

=> tan(α + β) = tanπ/2

hence, α + β = π/2


I HOPE IT'S HELP YOU DEAR,
THANKS