Question

☆☞ solve it
No spam ✖✖✖✖✖✖✖✖​

Answer 1

$\huge\bold\red{Answer}$

Given :-

$➡Tan A = a Tan B$

$➡Tan B = \frac{1}{b}$

$➡TanA$

$➡cot B = a/ Tan A$

( Equation 1 )

$➡sin A = b \: sin B$

$➡sin B = \frac{1}{b} b sin A$

$➡cosec B = b / Sin A$

( Equation 2 )

We know that , cosec ² B - cot² B = 1

From 1 and 2 Equation :-

$↪ \frac{ {b}^{2} } { \sin ^{2} A} - \frac{ {a}^{2} }{ { \tan }^{2}A} = 1$

$↪ \frac{ {b}^{2} }{ \sin ^{2} (A) } - \frac{ {a}^{2} \cos ^{2} (A ) }{ \sin ^{2} (A) } = 1$

$[ Where = \tan(A) = \frac{ \sin(A) }{ \cos(A) } ]$

$↪ \frac{ {b}^{2} - {a}^{2} \cos ^{2} (A) }{ \sin ^{2} (A) } = 1$

$= {b}^{2} - {a}^{2} \cos ^{2} (A) = \sin ^{2} (A)$

$= {b}^{2} - {a}^{2} \cos ^{2} (A) = 1 - \cos ^{2} (A)$

$= {b}^{2} - 1 = ( {a}^{2} - 1) \cos ^{2} A = \frac{ {b}^{2} - 1}{ {a}^{2} - 1 } = \cos ^{2} (A)$

Hence Proved ♠♠

Answer 2

Answer:

Given :-

➡Tan A = a Tan B➡TanA=aTanB

➡Tan B = \frac{1}{b}➡TanB=

b

1

➡TanA➡TanA

➡cot B = a/ Tan A➡cotB=a/TanA

( Equation 1 )

➡sin A = b \: sin B➡sinA=bsinB

➡sin B = \frac{1}{b} b sin A➡sinB=

b

1

bsinA

➡cosec B = b / Sin A➡cosecB=b/SinA

( Equation 2 )

We know that , cosec ² B - cot² B = 1

From 1 and 2 Equation :-

↪ \frac{ {b}^{2} } { \sin ^{2} A} - \frac{ {a}^{2} }{ { \tan }^{2}A} = 1↪

sin

2

A

b

2

−

tan

2

A

a

2

=1

↪ \frac{ {b}^{2} }{ \sin ^{2} (A) } - \frac{ {a}^{2} \cos ^{2} (A ) }{ \sin ^{2} (A) } = 1↪

sin

2

(A)

b

2

−

sin

2

(A)

a

2

cos

2

(A)

=1

[ Where = \tan(A) = \frac{ \sin(A) }{ \cos(A) } ][Where=tan(A)=

cos(A)

sin(A)

]

↪ \frac{ {b}^{2} - {a}^{2} \cos ^{2} (A) }{ \sin ^{2} (A) } = 1↪

sin

2

(A)

b

2

−a

2

cos

2

(A)

=1

= {b}^{2} - {a}^{2} \cos ^{2} (A) = \sin ^{2} (A)=b

2

−a

2

cos

2

(A)=sin

2

(A)

= {b}^{2} - {a}^{2} \cos ^{2} (A) = 1 - \cos ^{2} (A)=b

2

−a

2

cos

2

(A)=1−cos

2

(A)

= {b}^{2} - 1 = ( {a}^{2} - 1) \cos ^{2} A = \frac{ {b}^{2} - 1}{ {a}^{2} - 1 } = \cos ^{2} (A)=b

2

−1=(a

2

−1)cos

2

A=

a

2

−1

b

2

−1

=cos

2

(A)

Hence Proved ♠♠