Question

Sintheta +costheta=√3 then prove that tan theta +cot theta=1

Answer 1

sin +cos =√3
so ,by squaring both sides we get
= sin^2 + cos^2 + 2sincos=3
=1+2sincos=3
=2sincos=2
=sincos=1
tan + cot
= (sin/cos)+(cos/sin)
=(sin^2 +cos^2)/sincos
=1/1
=1
please mark as brainliest

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

sinθ + cosθ = √3 ... (1)

Now,

Here

Squaring on both sides

(sinθ + cosθ)² = (√3)²

Identity :-

${\boxed{\sf\:{(a+b)^2=a^2+b^2+2ab}}}$

Hence,

sin²θ + cos²θ + 2sinθcosθ = 3

As we know that :-

sin²θ + cos²θ = 1

Here, we get

2sinθcosθ = 2

sinθcosθ = 1..... (2)

Now

tanθ + cotθ

$\tt{\rightarrow\dfrac{sin\theta}{cos\theta}+\dfrac{cos\theta}{sin\theta}}$

$\tt{\rightarrow\dfrac{sin^2 \theta+cos^2 \theta}{(sin\theta)(cos\theta)}}$

$\tt{\rightarrow\dfrac{1}{(sin\theta)(cos\theta)}}$

= 1

Here we get :-

LHS = RHS

$\boxed{\begin{minipage}{11 cm} Fundamental Trignometric Indentities \\ \\ \sin^{2}\theta+\cos^{2}\theta =1 \\ \\ 1+tan^{2}\theta=\sec^{2}\theta \\ \\ 1 + cot^{2}\theta=\text{cosec}^2\theta \\ \\ tan\theta =\dfrac{sin\theta}{cos\theta} \\ \\ cot\theta =\dfrac{cos\theta}{sin\theta} \end{minipage}}$