Question

8. () If A + B = 90°, then express cos B in terms of simplest trigonometric ratio of 4
(ii) If X + Y = 90°, then express cos X in terms of simplest trigonometric ratio of Y
9. (1) If A + B = 90°, sin A = a, sin B = b, then prove that
Saya² + b 2 = 1
(b)/tan A=​

Answer 1

➦ Q.8

Given,A+B=90°

$⇒B=90°-A...(1)$

In equation (1) , multiplying both the sides by cos

$⇒ \cos(B) = \cos(90° - A)$

In this known, Sin θ = Cos θ (90°-θ)

$⇒ \cos(B) = \sin(A)$

Given ,X + Y=90°

$⇒ \cos X = \cos(90° -Y )$

In this known, Sin θ = Cos θ (90°-θ)

$⇒x = 90° - y...(1)$

In equation (1) , multiplying both the sides by cos

$⇒ \cos X = \sin Y$

➦ Q.9

GIVEN :- A + B = 90° , Sin A = Sin a , Sin B= Sin b

Now ,

$LHS =a² + b² = { \sin }^{2} A + { \sin }^{2} B$

$={ \sin }^{2} A + { \sin }^{2} (90°-A)$

$⇒{ \sin }^{2} A + { \cos }^{2} A$

∴{ A + B = 90°

∴B = 90° - A}

$∴1 = RHS$

➣Let's Recall

➛sin(90°−x) = cos x

➛cos(90°−x) = sin x

➛tan(90°−x) = cot x

➛cot(90°−x) = tan x

➛sec(90°−x) = csc x

➛csc(90°−x) = sec x

Answer 2

Answer:

Given,A+B=90°

⇒B=90°-A...(1)⇒B=90°−A...(1)

In equation (1) , multiplying both the sides by cos

⇒ \cos(B) = \cos(90° - A)⇒cos(B)=cos(90°−A)

In this known, Sin θ = Cos θ (90°-θ)

⇒ \cos(B) = \sin(A)⇒cos(B)=sin(A)

Given ,X + Y=90°

⇒ \cos X = \cos(90° -Y )⇒cosX=cos(90°−Y)

In this known, Sin θ = Cos θ (90°-θ)

⇒x = 90° - y...(1)⇒x=90°−y...(1)

In equation (1) , multiplying both the sides by cos

⇒ \cos X = \sin Y⇒cosX=sinY

➦ Q.9

GIVEN :- A + B = 90° , Sin A = Sin a , Sin B= Sin b

Now ,

LHS =a² + b² = { \sin }^{2} A + { \sin }^{2} BLHS=a²+b²=sin

2

A+sin

2

B

={ \sin }^{2} A + { \sin }^{2} (90°-A)=sin

2

A+sin

2

(90°−A)

⇒{ \sin }^{2} A + { \cos }^{2} A⇒sin

2

A+cos

2

A

∴{ A + B = 90°

∴B = 90° - A}

∴1 = RHS∴1=RHS

➣Let's Recall

➛sin(90°−x) = cos x

➛cos(90°−x) = sin x

➛tan(90°−x) = cot x

➛cot(90°−x) = tan x

➛sec(90°−x) = csc x

➛csc(90°−x) = sec x