Question

sin A + cosec a square + Cos A + sec a square is equal to 7 + tan square A + cot square A

​

Answer 1

Appropriate Question :-

Prove that

$\rm \: {(sinA + cosec)}^{2} + {(cosA + secA)}^{2} = 7 + {tan}^{2}A + {cot}^{2}A$

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

Consider LHS

$\rm :\longmapsto\: {(sinA + cosec)}^{2} + {(cosA + secA)}^{2}$

Using this identity,

$\underbrace{ \boxed{ \bf \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

we get

$\rm \:= {sin}^{2}A + {cosec}^{2}A + 2sinAcosecA + {cos}^{2}A + {sec}^{2}A + 2cosAsecA$

We know,

$\underbrace{ \boxed{ \bf \: cosecA = \frac{1}{sinA}}} \: \: \: and \: \: \: \underbrace{ \boxed{ \bf \: secA = \frac{1}{cosA}}}$

Using these, we get

$\rm \:=( {sin}^{2}A + {cos}^{2}A) + 2sinA \times \dfrac{1}{cosecA} +{cosec}^{2}A + {sec}^{2}A + 2cosA \times \dfrac{1}{cosA}$

We know,

$\underbrace{ \boxed{ \bf \: {sin}^{2}A + {cos}^{2}A = 1}} \: \: \underbrace{ \boxed{ \bf \: {sec}^{2}A = 1 + {tan}^{2}A}}$

and

$\underbrace{ \boxed{ \bf \: {cosec}^{2} A = 1 + {cot}^{2}A}}$

So, on using these Identities, we get

$\rm \: = \: 1 + 2 + (1 + {cot}^{2}A) + (1 + {tan}^{2}A) + 2$

$\rm \: = \: 7 + {tan}^{2}A + {cot}^{2}A$

= RHS

Hence, Proved

Thus,

$\boxed{\rm \: {(sinA + cosec)}^{2} + {(cosA + secA)}^{2} = 7 + {tan}^{2}A + {cot}^{2}A}$

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