Question

*9. यदि tan 15° = 2-✓3 है तो tan 75° का मान ज्ञात कीजिए। मात जात कीजिए​

Answer 1

Step-by-step explanation:

we have,

$\tt \tan 15^{\circ}=2-\sqrt{3}$

$\tt \boxed{\tan (A+B)= \dfrac{\tan A+ \tan B}{1- \tan A \tan B}}$

$\tt \tan 75 = \tan (60+15)$

$\tt \tan 60^{\circ} = \sqrt{3}$

$\tt \implies \tan (60+15)= \dfrac{\tan 60+ \tan 15}{1- \tan 60 \tan 15}$

$\tt \implies \tan (75)= \dfrac{\sqrt{3}+ 2-\sqrt{3}}{1- \sqrt{3} \times (2-\sqrt{3}) }$

$\tt \implies \tan (75)= \dfrac{\cancel{\sqrt{3}}+ 2-\cancel{\sqrt{3}}}{1- 2\sqrt{3}-3 }$

$\tt \implies \tan (75)= \dfrac{2}{-2-2\sqrt{3}}$

Answer 2

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

Given that

$\rm :\longmapsto\:tan15 \degree = 2 - \sqrt{3}$

Now, Consider

$\red{\rm :\longmapsto\:tan75\degree}$

can be rewritten as

$\rm \:  =  \: tan(90\degree - 15\degree)$

We know,

$\boxed{ \tt{ \: tan(90\degree - x) = cotx \: }}$

So, using this, we get

$\rm \:  =  \: cot15\degree$

$\rm \:  =  \: \dfrac{1}{tan15\degree}$

$\rm \:  =  \: \dfrac{1}{2 - \sqrt{3} }$

$\rm \:  =  \: \dfrac{1}{2 - \sqrt{3} } \times \dfrac{2 + \sqrt{3} }{2 - \sqrt{3} }$

$\rm \:  =  \: \dfrac{2 + \sqrt{3} }{ {(2)}^{2} - {( \sqrt{3}) }^{2} }$

$\rm \:  =  \: \dfrac{2 + \sqrt{3} }{4 - 3}$

$\rm \:  =  \: \dfrac{2 + \sqrt{3} }{1}$

$\rm \:  =  \: 2 + \sqrt{3}$

Hence,

$\rm \implies\:\boxed{ \tt{ \: tan75\degree =  \: 2 + \sqrt{3} \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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