Question

Answer the above file​

Answer 1

Refer to the attachment for complete solution.

All Trigonometric Basic Formulas :-

• sin A = Perpendicular / Hypotenuse
• cos A = Base / Hypotenuse
• tan A = Perpendicular / base
• cosec A = Hypotenuse / Perpendicular
• cos A = Hypotenuse / base
• cot A = Base / Hypotenuse
• sec² A - tan² A = 1
• 1 + cot² A = cosec²A
• sin² A + cos² A = 1
• sin A = 1 / cosec A
• cos A = 1 / sec A
• tan A = 1 / cot A
• cosec A = 1 / sin A
• sec A = 1 / cos A
• cot A = 1 / tan A
• tan A = sin A / cos A
• tan A = sec A / cosec A
• cot A = cos A / sin A
• cot A = cosec A / cos A
• sin ( 90° - A ) = cos A
• cos ( 90° - A ) = sin A
• tan ( 90° - A ) = cot A
• cot ( 90° - A ) = tan A
• cosec ( 90° - A ) = sec A
• sec ( 90° - A ) = cos A

Some Commonly used identities :-

• ( A + B ) ( A - B ) = A² - B²
• ( A + B )² = A² + B² + 2 AB
• ( A - B )² = A² + B² - 2 Ab
• ( A + B + C)² = A² + B² + C² + 2 AB + 2 BC + 2 CA
• A = √[ ( A ) ²]
• √ A² = ± A

Answer 2

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

Prove that,

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

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

Consider LHS

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

can be rewritten as

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

On rationalizing the denominator, we get

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

We know,

$\boxed{ \rm \:(x + y)(x - y) = {x}^{2} - {y}^{2}} \\ \\ \boxed{ \rm \:{(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}$

So, using these Identities,

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

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

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

We know,

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

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

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

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

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

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

$\rm \:  =  \:\dfrac{1 - cosA}{sinA}$

can be further rewritten as

$\rm \:  =  \: \sqrt{\dfrac{ {(1 - cosA)}^{2} }{ {sin}^{2} A} }$

$\rm \:  =  \: \sqrt{\dfrac{ {(1 - cosA)}^{2} }{ 1 - {cos}^{2} A} }$

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

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

Hence,

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

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