Question

if cosecA- cot A = 1/3 then the value of ( cosec A + CotA ) ​

Answer 1

Question:

If $cosec A-cot A=1/3$ then the value of $(cosec A +cot A)$

Answer:

$cosec A+ cot A=3$

Step-by-step explanation:

$Given,$

$Cosec A-cot A=\frac{1}{3}$

$⇒(cosec A-cot A)(cosec A+ cot A)=\frac{1}{3}(cosec A+ cot A)$

$⇒(cosec^{2}A-cot^{2}A=\frac{1}{3}(cosec A + cot A)$

$⇒1=\frac{1}{3}(cosec A + cot A)$

$⇒\boxed{cosec^{2}A-cot^{2}A=1}$

$⇒3=cosec A + cot A$

$therefore,$

If $cosec A - cot A=\frac{1}{2}$

$then,$

$cosec A + cot A=3$

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Answer 2

$\bigstar \underline{\underline{\sf \bf Given:-}}$

• Cosec A -Cot A = 1/3

$\bigstar \underline{\underline{\sf \bf To\ Find:-}}$

• Value of Cosec A + Cot A .

$\bigstar \underline{\underline{\sf \bf Solution:-}}$

We need to know :

------------------------------

❥ Cosec A = 1/sin A

❥ Cot A = Cos A / Sin A

❥ 1-cos²A = Sin²A

--------------------------------

Now,

Given Cosec A - cot A = 1/3

$:\implies \sf \frac{1}{sinA} - \frac{cosA}{sinA} =\frac{1}{3} \\\\\\:\implies \sf \frac{1-cosA}{sinA} =\frac{1}{3} \\\\\\\sf Squaring\ on\ both\ sides,\\\\\\:\implies \sf \frac{(1-cosA)^2}{sin^2A} = \frac{1}{9} \\\\\\:\implies \sf \frac{(1-cosA)^2}{1-cos^2A} =\frac{1}{9}\\\\\\:\implies \sf \frac{(1-cosA)^2}{(1+cosA)(1-cosA)} \\\\\\:\implies \sf \frac{1-cosA}{1+cosA} =\frac{1}{9}$

Multiply Numerator and denominator of LHS with (1+cosA).

$:\implies \sf \frac{(1-cosA)(1+cosA)}{(1+cosA)(1+cosA)} =\frac{1}{9} \\\\\\:\implies \sf \frac{1-cos^2 A}{(1+cosA)^2} = \frac{1}{9}\\\\\\:\implies \sf \frac{sin^2A}{(1+cosA)^2} =\frac{1}{9}$

Applying square root on both sides we get :

$:\implies \sf \frac{sinA}{1+cosA} =\frac{1}{3}\\\\\\:\implies \sf \frac{1+cosA}{sinA} =\frac{3}{1}\\\\\\:\implies \sf \frac{1}{sinA} +\frac{cosA}{sinA} =\frac{3}{1}\\\\\\:\implies \boxed{\boxed{\pink{\sf CosecA+CotA=3}}}$

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