Question

if x= a sec tita + b tan tita and y = a tan tita + b sec tita , then prove that x²-y²=a²-b²​

Answer 1

$\star\:\:\:\sf\large\underline\blue{Answer:-}$

Given,

$\sf x = a \sec(x) + b \tan(x)$

$\sf y = a \tan(x) + b \sec(x)$

To prove:-

$\sf {x}^{2} - {y}^{2} = {a}^{2} - {b}^{2}$

Proof:-

$\sf {x}^{2} - {y}^{2}$

$\sf = {(a \sec(x) + b \tan(x)) }^{2} - {(a \tan(x) + b \sec(x) )}^{2}$

$\sf = {a}^{2} { \sec }^{2} (x) + {b}^{2} \tan^{2}(x) + 2ab \sec(x) \tan(x ) - {a}^{2} { \tan}^{2} (x) - {b}^{2} { \sec }^{2} (x) - 2ab \sec(x) \tan(x)$

$\sf = {a}^{2} ( { \sec }^{2} (x) - \tan ^{2} (x) ) + {b}^{2} ( \tan ^{2} (x) - \sec^{2} (x) )$

$\sf {a}^{2} \times 1 - {b}^{2} ( \sec ^{2} (x) - \tan ^{2} (x) )$

$\sf= {a}^{2} - {b}^{2} \times 1$

$\sf = {a}^{2} - {b}^{2}$

Therefore,

$\boxed{ \sf \large {x}^{2} - {y}^{2} = {a}^{2} - {b}^{2} }$

Hence Proved.

Answer 2

Answer:

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:−

,

\ = \() + \()=()+()

\ = \() + \()=()+()

:-

\ {}^{2} - {}^{2} = {}^{2} - {}^{2}

2

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2

=

2

−

2

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\ {}^{2} - {}^{2}

2

−

2

\ = {( \() + \()) }^{2} - {( \() + \() )}^{2}=(()+())

2

−(()+())

2

\ = {}^{2} { \ }^{2} () + {}^{2} \^{2}() + 2 \() \( ) - {}^{2} { \}^{2} () - {}^{2} { \ }^{2} () - 2 \() \()=

2

2

()+

2

2

()+2()()−

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()−

2

2

()−2()()

\ = {}^{2} ( { \ }^{2} () - \ ^{2} () ) + {}^{2} ( \ ^{2} () - \^{2} () )=

2

(

2

()−

2

())+

2

(

2

()−

2

())

\ {}^{2} \ 1 - {}^{2} ( \ ^{2} () - \ ^{2} () )

2

×1−

2

(

2

()−

2

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\= {}^{2} - {}^{2} \ 1=

2

−

2

×1

\ = {}^{2} - {}^{2}=

2

−

2

,

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−

2

=

2

−

2

.