Question

help me in solving this iam a teacher of class 10​

Answer 1

Topic

• Trigonometry

A branch of Mathematics about study of triangles.

• Trigonometric Identity

When an equation is always true we call it an identity.

$\implies \boxed{\sin^{2} x+\cos^{2} x=1}$

Solution

Modifying the basic identity, we get the following.

$\implies \sin^{2} x+\cos^{2} x=1$

$\implies 1-\cos^{2} x=\sin^{2} x$

$\implies \dfrac{1}{\sin^{2} x} -\dfrac{1}{\tan^{2} x} =1$

$\implies \boxed{\csc^{2} x-\cot^{2} x=1}$

The given equation is equivalent to the following.

$\implies \dfrac{1}{\csc x+\cot x} +\dfrac{1}{\csc x-\cot x} =\dfrac{2}{\sin x}$

It is enough to prove this equation is true.

Left-hand Side

$=\dfrac{\csc x-\cot x+\csc x+\cot x}{\csc^{2} x-\cot^{2} x}$

$=\dfrac{2\csc x}{\csc^{2} x-\cot^{2} x}$

$=\dfrac{2\csc x}{1}$

$=\dfrac{2}{\sin x}$

= Right-hand Side

This is the required answer.

Answer 2

$\large\underline{\sf{Given \:Question - }}$

Prove that

$\rm \: \dfrac{1}{cosecx + cotx} - \dfrac{1}{sinx} = \dfrac{1}{sinx} - \dfrac{1}{cosecx - cotx}$

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm :\longmapsto\:\dfrac{1}{cosecx + cotx} - \dfrac{1}{sinx}$

On rationalizing the first term, we get

$\rm \:  =  \:  \: \dfrac{1}{cosecx + cotx} \times \dfrac{cosecx - cotx}{cosecx - cotx} - cosecx$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: \dfrac{1}{sinx} = cosecx\bigg \}}$

$\rm \:  =  \:  \: \dfrac{cosecx - cotx}{ {cosec}^{2} x - {cot}^{2}x } - cosecx$

$\: \: \: \: \: \: \red{\bigg \{ \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}$

$\rm \:  =  \:  \: cosecx - cotx - cosecx$

$\red{\bigg \{ \because \: {cosec}^{2}x - {cot}^{2}x = 1 \bigg \}}$

$\rm \:  =  \:  \: - cotx$

Hence,

$\bf :\longmapsto\:\dfrac{1}{cosecx + cotx} - \dfrac{1}{sinx} = - \: cotx - - (1)$

Consider RHS

$\rm :\longmapsto\:\dfrac{1}{sinx} - \dfrac{1}{cosecx + cotx}$

On rationalizing the second term, we get

$\rm \:  =  \:  \: cosecx - \dfrac{1}{cosecx - cotx} \times \dfrac{cosecx + cotx}{cosecx + cotx}$

$\rm \:  =  \:  \: cosecx - \dfrac{cosecx + cotx}{ {cosec}^{2}x - {cot}^{2}x }$

$\rm \:  =  \:  \: cosecx - (cosecx + cotx)$

$\rm \:  =  \:  \: cosecx - cosecx - cotx$

$\rm \:  =  \:  \: - \: cotx$

Hence,

$\bf :\longmapsto\:\dfrac{1}{sinx} - \dfrac{1}{cosecx + cotx} = - \: cotx - - (2)$

From equation (1) and equation (2), we get

$\bf \: \dfrac{1}{cosecx + cotx} - \dfrac{1}{sinx} = \dfrac{1}{sinx} - \dfrac{1}{cosecx - cotx}$

Hence, Proved

Additional Information :

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1