Question

If cosecA=2m and cot A=2/m,then 2(m²-1/m²)=? ​

Answer 1

Answer:

$\huge{\underline{\mathtt{\red{❥A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}$It is given that cosec{\theta}+cot{\theta}=mcosecθ+cotθ=m , then

\frac{1}{sin{\theta}}+\frac{cos{\theta}}{sin{\theta}}=m

sinθ1 + sinθcosθ =m

\frac{1+cos{\theta}}{sin{\theta}}=m

sinθ

1+cosθ =m

(\frac{1+cos{\theta}}{sin{\theta}})^2=m^2( sinθ

1+cosθ ) 2 =m2

m^2=\frac{1+cos^2{\theta}+2cos{\theta}}{sin^2{\theta}}m

2 =sin 2θ

1+cos 2θ+2cosθ

Now, \frac{m^2-1}{m^2+1}=\frac{1+cos^2{\theta}+2cos{\theta}-sin^2{\theta}}{1+cos^2{\theta}+2cos{\theta}+sin^2{\theta}}

m 2 +1

m 2 −1= 1+cos 2

θ+2cosθ+sin

2 θ1+cos 2 θ+2cosθ−sin2θ

\frac{m^2-1}{m^2+1}=\frac{2cos^2{\theta}+2cos{\theta}}{2+2cos{\theta}}

m 2+1m 2−1=

2+2cosθ

2cos 2 θ+2cosθ

\frac{m^2-1}{m^2+1}=cos{\theta}

m 2+1m 2−1 =cosθ

Hence proved.

Answer 2

Answer:

Let, cosecA+cotA=m

Then 

m2−1m2+1

=cosec2A+cot2A+2cosecA.cotA−1cosec2A+cot2A+2cosecA.cotA+1

=2cot2A+2cosecA.cotA2cosec2A+2cosecA.cotA [Since cosec2A−1=cot2A and 1+cot2A=cosec2A]

=cotAcosecA=secA.........(1).

Now 

m2+1m2−1=cosA [Using (1)]

Step-by-step explanation:

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