Question

Q4. If sin A/sin B = √3/2 and cos A/cos B = √5/2, 0 < A, B < π/2, then find the value of tan A + tan B.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:\dfrac{sinA}{sinB} = \dfrac{ \sqrt{3} }{2}$

$\rm :\longmapsto\: sinA= \dfrac{ \sqrt{3} }{2} sinB - - - (1)$

Also,

Given that

$\rm :\longmapsto\:\dfrac{cosA}{cosB} = \dfrac{ \sqrt{5} }{2}$

$\rm :\longmapsto\:cosA = \dfrac{ \sqrt{5} }{2} cosB - - - (2)$

On squaring equation (1) and (2) and adding, we get

$\rm :\longmapsto\: {sin}^{2}A +{cos}^{2}A = \dfrac{3}{4} {sin}^{2}B + \dfrac{5}{4} {cos}^{2}B$

$\rm :\longmapsto\: 1 = \dfrac{3}{4} {sin}^{2}B + \dfrac{5}{4} {cos}^{2}B$

$\: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: {sin}^{2}x + {cos}^{2} x = 1 \bigg \}}$

$\rm :\longmapsto\: {3sin}^{2}B + {5cos}^{2}B = 4$

$\rm :\longmapsto\: {3sin}^{2}B + {3cos}^{2}B + {2cos}^{2}B = 4$

$\rm :\longmapsto\: 3({sin}^{2}B + {cos}^{2}B) + {2cos}^{2}B = 4$

$\rm :\longmapsto\: 3(1) + {2cos}^{2}B = 4$

$\: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: {sin}^{2}x + {cos}^{2} x = 1 \bigg \}}$

$\rm :\longmapsto\: 3 + {2cos}^{2}B = 4$

$\rm :\longmapsto\: {2cos}^{2}B = 4 - 3$

$\rm :\longmapsto\: {2cos}^{2}B = 1$

$\rm :\longmapsto\: {cos}^{2}B = \dfrac{1}{2} - - - - (3)$

$\rm :\longmapsto\:sec^{2}B = 2$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: secx = \dfrac{1}{cosx} \bigg \}}$

$\rm :\longmapsto\:1 + {tan}^{2}B = 2$

$\rm :\longmapsto\:{tan}^{2}B = 2 - 1$

$\rm :\longmapsto\:{tan}^{2}B = 1$

$\bf\implies \:tanB = 1 - - - (4)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: 0 < B < \dfrac{\pi}{2} \bigg \}}$

Now,

From equation (2), we have

$\rm :\longmapsto\:cosA = \dfrac{ \sqrt{5} }{2} cosB$

Squaring both sides we get

$\rm :\longmapsto\:cos^{2} A = \dfrac{5}{4} cos^{2} B$

$\rm :\longmapsto\: {cos}^{2}A = \dfrac{5}{4} \times \dfrac{1}{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: {cos}^{2}B = \dfrac{1}{2} \bigg \}}$

$\rm :\longmapsto\: {cos}^{2}A = \dfrac{5}{8}$

$\rm :\longmapsto\: {sec}^{2}A = \dfrac{8}{5}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: secx = \dfrac{1}{cosx} \bigg \}}$

$\rm :\longmapsto\: 1 + {tan}^{2}A = \dfrac{8}{5}$

$\rm :\longmapsto\:{tan}^{2}A = \dfrac{8}{5} - 1$

$\rm :\longmapsto\:{tan}^{2}A = \dfrac{8 - 5}{5}$

$\rm :\longmapsto\:{tan}^{2}A = \dfrac{3}{5}$

$\rm :\longmapsto\:{tan}A = \dfrac{ \sqrt{3} }{ \sqrt{5} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: 0 < A < \dfrac{\pi}{2} \bigg \}}$

$\rm :\longmapsto\:{tan}A = \dfrac{ \sqrt{3} }{ \sqrt{5} } \times \dfrac{ \sqrt{5} }{ \sqrt{5} }$

$\bf :\longmapsto\:tanA = \dfrac{ \sqrt{15} }{5}$

Hence,

$\bf\implies \:tanA \: + \: tanB = \dfrac{ \sqrt{15} }{5} + 1$

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1