Question

Sin theta + cos theta = √ 2 then (tan theta + cot theta) = ?​

Answer 1

Step-by-step explanation:

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Answer 2

$\huge{\mathbb{\red{ANSWER:-}}}$

$\sf{SinO + CosO =\sqrt{2}}$

$\sf{SinO(1+CotO)=\sqrt{2}}$

$\sf{(1 + CotO)=\sqrt{2}CosecO}$

$\sf{Doing \: square \: both \: side-}$

$\sf{(1+CotO)^2 = 2Cosec^2 O}$

$\sf{(1+Cot^2 O + 2CotO)=2Cosec^2 O}$

$\sf{Cosec^2 O + 2CotO = 2Cosec^2 O}$

$\sf{Cosec^2 O = 2CotO}$

$\sf{\dfrac{1}{Sin^2 O}=2\dfrac{CosO}{SinO}}$

$\sf{\dfrac{1}{SinO}=2CosO}$

$\sf{2SinOCosO=1}$

$\sf{\dfrac{1}{SinOCosO}=2}$ --(1)

$\sf{Now}$

$\sf{tanO + CotO}$

$\sf{\dfrac{SinO}{CosO}+\dfrac{CosO}{SinO}}$

$\sf{\dfrac{(Sin^2 O + Cos^2 O)}{SinOCosO}}$

$\sf{\dfrac{1}{SinOCosO}}$

$\sf{From \: eq.(1)}$

$=\sf{2}$

$\sf{(tanO + CotO) = 2}$

Using properties :-

1)$\sf{Cosec^2 O = 1 + Cot^2 O}$

2)$\sf{Sin^2 O + Cos^2 O = 1}$

3)$\sf{CosecO=\dfrac{1}{SinO}}$