Question

if cosec theta+cot theta=m then prove that (m square +1) cos theta=m square-1​

Answer 1

Step-by-step explanation:

Answer:-

$\huge{\rm{\pink{\underline{Given:-}}}}$

$\sf{Cosec \theta + Cot \theta = m}$

$\huge{\rm{\pink{\underline{To \ prove:-}}}}$

$\sf{(m^2+1)Cos \theta = m^2 -1}$

Concept:-

Trigonometry and its applications.

Let's Do!

$\sf{(m^2+1)Cos \theta = m^2 -1}$

$\sf{Cos \theta = \dfrac{m^2 -1}{(m^2+1)}}$ (1)

$\sf{Cosec \theta + Cot \theta = m}$

$\sf{\dfrac{1}{sin \theta} + \dfrac{cos \theta }{sin \theta} = m}$

$\sf{\dfrac{1 + cos \theta}{sin \theta} = m}$ (2)

Placing value of (2) in (1),

Numerator:-

$\sf{(\dfrac{1+ cos \theta}{sin \theta})^2 - 1 }$

$\sf{\dfrac{(1+ cos \theta)^2}{sin^2 \theta} - 1 }$

$\sf{\dfrac{(1 + cos^2 \theta+2cos \theta)}{sin^2 \theta} - 1 }$

$\sf{\dfrac{1 + cos^2 \theta+2cos \theta - sin^2 \theta}{sin^2 \theta}}$

$\sf{\dfrac{cos^2 \theta + cos^2 \theta+2cos \theta }{sin^2 \theta}}$

$\sf{\dfrac{2cos^2 \theta +2cos \theta }{sin^2 \theta}}$

Denominator:-

$\sf{(\dfrac{1+ cos \theta}{sin \theta})^2 + 1 }$

$\sf{\dfrac{(1+ cos \theta)^2}{sin^2 \theta} + 1 }$

$\sf{\dfrac{(1 + cos^2 \theta+2cos \theta)}{sin^2 \theta} + 1 }$

$\sf{\dfrac{1 + cos^2 \theta+2cos \theta + sin^2 \theta}{sin^2 \theta}}$

$\sf{\dfrac{1 +1+2cos \theta}{sin^2 \theta}}$

$\sf{\dfrac{2+2cos \theta}{sin^2 \theta}}$

Now, combining both:-

$\sf{\dfrac{\dfrac{2cos^2 \theta +2cos \theta }{sin^2 \theta}}{\dfrac{2+2cos \theta}{sin^2 \theta}}}$

Now, sin² theta gets cancelled.

$\sf{\dfrac{2 cos^2 \theta + 2 cos \theta }{2+2 cos \theta}}$

$\sf{\dfrac{2 cos \theta( cos \theta +1)}{2(1+ cos \theta)}}$

2 gets cancelled, 1 + cos theta cancelled, so left is cos theta.

Hence Proved!

Answer 2

$\huge\sf\underline{\pink{Answer:-}}$

$\large\sf{\orange{Given:-}}$

$\sf{\green{Cosec \theta + Cot \theta = m}}$

$\large\sf{\color{navy}{To \ prove:-}}$

$\sf{\green{(m^2+1)Cos \theta = m^2 -1}}$

$\large\sf{\purple{Concept:-}}$

Trigonometry and its applications.

$\large\sf{\color{gold}{Let's ~Do!}}$

$\sf{(m^2+1)Cos \theta = m^2 -1} \\ \\ </p><p>\sf{Cos \theta = \dfrac{m^2 -1}{(m^2+1)}} \\ \\ </p><p>\sf{Cosec \theta + Cot \theta = m} \\ \\ </p><p>\sf{\dfrac{1}{sin \theta} + \dfrac{cos \theta }{sin \theta} = m} \\ \\ </p><p>\sf{\dfrac{1 + cos \theta}{sin \theta} = m}$

Placing value of (2) in (1),

$\large \sf{ \underline{ \underline{ \red{Numerator:-}}}}$

$\sf{(\dfrac{1+ cos \theta}{sin \theta})^2 - 1 } \\ \\ </p><p>\sf{\dfrac{(1+ cos \theta)^2}{sin^2 \theta} - 1 } \\ \\ </p><p>\sf{\dfrac{(1 + cos^2 \theta+2cos \theta)}{sin^2 \theta} - 1 } \\ \\ \sf{\dfrac{1 + cos^2 \theta+2cos \theta - sin^2 \theta}{sin^2 \theta}} \\ \\ \sf{\dfrac{cos^2 \theta + cos^2 \theta+2cos \theta }{sin^2 \theta}} \\ \\ \sf{\dfrac{2cos^2 \theta +2cos \theta }{sin^2 \theta}}$

$\large \sf{ \underline{ \underline{ \red{Denominator:-}}}}$

$\sf{(\dfrac{1+ cos \theta}{sin \theta})^2 + 1 } \\ \\ </p><p>\sf{\dfrac{(1+ cos \theta)^2}{sin^2 \theta} + 1 } \\ \\ \sf{\dfrac{(1 + cos^2 \theta+2cos \theta)}{sin^2 \theta} + 1 } \\ \\ </p><p>\sf{\dfrac{1 + cos^2 \theta+2cos \theta + sin^2 \theta}{sin^2 \theta}} \\ \\ </p><p>\sf{\dfrac{1 +1+2cos \theta}{sin^2 \theta}} \\ \\ </p><p>\sf{\dfrac{2+2cos \theta}{sin^2 \theta}}$

$\large \sf{ \underline{ \color{midnightblue}{Now, combining \: both:-}}}$

$\sf{\dfrac{\dfrac{2cos^2 \theta +2cos \theta }{sin^2 \theta}}{\dfrac{2+2cos \theta}{sin^2 \theta}}}$

$\sf{ \color{purple}{Now, \: sin² \: theta \: gets \: cancelled.}}$

$\sf{\dfrac{2 cos^2 \theta + 2 cos \theta }{2+2 cos \theta}} \\ \\ </p><p>\sf{\dfrac{2 cos \theta( cos \theta +1)}{2(1+ cos \theta)}}$

2 gets cancelled, 1 + cos theta cancelled, so left is cos theta.