help me in solving this iam a teacher of class 10
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Topic
A branch of Mathematics about study of triangles.
When an equation is always true we call it an identity.
[tex]\implies \boxed{\sin^{2} x+\cos^{2} x=1}[/tex]
Solution
Modifying the basic identity, we get the following.
[tex]\implies \sin^{2} x+\cos^{2} x=1[/tex]
[tex]\implies 1-\cos^{2} x=\sin^{2} x[/tex]
[tex]\implies \dfrac{1}{\sin^{2} x} -\dfrac{1}{\tan^{2} x} =1[/tex]
[tex]\implies \boxed{\csc^{2} x-\cot^{2} x=1}[/tex]
The given equation is equivalent to the following.
[tex]\implies \dfrac{1}{\csc x+\cot x} +\dfrac{1}{\csc x-\cot x} =\dfrac{2}{\sin x}[/tex]
It is enough to prove this equation is true.
Left-hand Side
[tex]=\dfrac{\csc x-\cot x+\csc x+\cot x}{\csc^{2} x-\cot^{2} x}[/tex]
[tex]=\dfrac{2\csc x}{\csc^{2} x-\cot^{2} x}[/tex]
[tex]=\dfrac{2\csc x}{1}[/tex]
[tex]=\dfrac{2}{\sin x}[/tex]
= Right-hand Side
This is the required answer.
[tex]\large\underline{\sf{Given \:Question - }}[/tex]
Prove that
[tex] \rm \: \dfrac{1}{cosecx + cotx} - \dfrac{1}{sinx} = \dfrac{1}{sinx} - \dfrac{1}{cosecx - cotx} [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Consider LHS
[tex]\rm :\longmapsto\:\dfrac{1}{cosecx + cotx} - \dfrac{1}{sinx} [/tex]
On rationalizing the first term, we get
[tex]\rm \: = \: \: \dfrac{1}{cosecx + cotx} \times \dfrac{cosecx - cotx}{cosecx - cotx} - cosecx[/tex]
[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: \dfrac{1}{sinx} = cosecx\bigg \}}[/tex]
[tex]\rm \: = \: \: \dfrac{cosecx - cotx}{ {cosec}^{2} x - {cot}^{2}x } - cosecx[/tex]
[tex] \: \: \: \: \: \: \red{\bigg \{ \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}[/tex]
[tex]\rm \: = \: \: cosecx - cotx - cosecx[/tex]
[tex]\red{\bigg \{ \because \: {cosec}^{2}x - {cot}^{2}x = 1 \bigg \}}[/tex]
[tex]\rm \: = \: \: - cotx[/tex]
Hence,
[tex]\bf :\longmapsto\:\dfrac{1}{cosecx + cotx} - \dfrac{1}{sinx} = - \: cotx - - (1)[/tex]
Consider RHS
[tex]\rm :\longmapsto\:\dfrac{1}{sinx} - \dfrac{1}{cosecx + cotx} [/tex]
On rationalizing the second term, we get
[tex]\rm \: = \: \: cosecx - \dfrac{1}{cosecx - cotx} \times \dfrac{cosecx + cotx}{cosecx + cotx} [/tex]
[tex]\rm \: = \: \: cosecx - \dfrac{cosecx + cotx}{ {cosec}^{2}x - {cot}^{2}x } [/tex]
[tex]\rm \: = \: \: cosecx - (cosecx + cotx)[/tex]
[tex]\rm \: = \: \: cosecx - cosecx - cotx[/tex]
[tex]\rm \: = \: \: - \: cotx[/tex]
Hence,
[tex]\bf :\longmapsto\:\dfrac{1}{sinx} - \dfrac{1}{cosecx + cotx} = - \: cotx - - (2)[/tex]
From equation (1) and equation (2), we get
[tex] \bf \: \dfrac{1}{cosecx + cotx} - \dfrac{1}{sinx} = \dfrac{1}{sinx} - \dfrac{1}{cosecx - cotx} [/tex]
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