Question

Prove that sinA/secA+tanA-1+cosA/cosecA +cotA-1=1​

Answer 1

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

Consider LHS

$\rm :\longmapsto\:\dfrac{sinA}{secA + tanA - 1} + \dfrac{cosA}{cosecA + cotA - 1}$

We know,

$\boxed{ \tt{ \: secx = \frac{1}{cosx} \: }}$

$\boxed{ \tt{ \: tanx = \frac{sinx}{cosx} \: }}$

$\boxed{ \tt{ \: cosecx = \frac{1}{sinx} \: }}$

$\boxed{ \tt{ \: cotx = \frac{cosx}{sinx} \: }}$

So, on substituting all these values, we get

$\rm \: = \:\dfrac{sinA}{\dfrac{1}{cosA} + \dfrac{sinA}{cosA} - 1} + \dfrac{cosA}{\dfrac{1}{sinA} + \dfrac{cosA}{sinA} - 1}$

$\rm \: = \:\dfrac{sinA}{\dfrac{1 + sinA - cosA}{cosA} } + \dfrac{cosA}{\dfrac{1 + cosA - sinA}{sinA}}$

$\rm \: = \:\dfrac{sinAcosA}{1 + sinA - cosA} + \dfrac{sinAcosA}{1 + cosA - sinA}$

can be further rewritten as

$\rm= \:sinAcosA\bigg[\dfrac{1}{1 + sinA - cosA}+\dfrac{1}{cosA - sinA} \bigg]$

$\rm = sinAcosA\bigg[\dfrac{1 + cosA - sinA + 1 + sinA - cosA}{(1 + sinA - cosA)(1 + cosA - sinA)} \bigg]$

$\rm = sinAcosA\bigg[\dfrac{2}{(1 + sinA - cosA)(1 + cosA - sinA)} \bigg]$

can be further rewritten as

$\rm = sinAcosA\bigg[\dfrac{2}{[1 +(sinA - cosA)][(1 - [sinA - cosA)]} \bigg]$

$\rm \: = \:\dfrac{2 \: sinA \: cosA}{ {1}^{2} - {(sinA - cosA)}^{2} }$

$\rm \: = \:\dfrac{2 \: sinA \: cosA}{1 - ( {sin}^{2}A + {cos}^{2}A - 2sinAcosA) }$

$\rm \: = \:\dfrac{2 \: sinA \: cosA}{1 - ( 1 - 2sinAcosA) }$

$\rm \: = \:\dfrac{2 \: sinA \: cosA}{1 - 1 + 2sinAcosA}$

$\rm \: = \:\dfrac{2 \: sinA \: cosA}{2sinAcosA}$

$\rm \: = \:1$

Hence,

$\red{\boxed{ \sf{ \: \:\dfrac{sinA}{secA + tanA - 1} + \dfrac{cosA}{cosecA + cotA - 1} = 1}}}$

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

Answer 2

Answer:

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