Question

plz help me ....irrevalent answers will be reported answer it correctly.​

Answer 1

Solution

Given :-

• sec A - tan A = x

To Prove :-

• (1+x²)/(1-x²) = cosec A

Prove

Using Formula

★ sec² A - tan² A = 1

★ tan A = sin A/cos A

★ cos A = 1/sec A

________________

So, First take L.H.S.

➡ (1+x²)/(1-x²)

Keep value of x .

➡ [ 1 + (sec A - tan A)²]/[1 - (sec A - tan A)²]

➡[1+(sec² A + tan² A - 2 tan A sec A)]/[1-(sec² A + tan² A - 2 tan A sec A)]

➡[1+sec² A + tan² A - 2tan A . sec A]/[1-sec² A - tan² A + 2 tan A sec A]

➡[(sec² A - tan² A)+(sec² A + tan² A - 2tan A . sec A)]/[(sec² A - tan² A)-sec² A - tan² A + 2 tan A sec A]

➡2(sec² A - sec A tan A)/2(-tan² A - sec A tan A)

➡sec A(sec A - tan A)/tan A(sec A - tan A)

➡sec A / tan A

➡ (1/cos A )/( sin A/cos A)

➡1/cos A × cos A /sin A

➡1/sin A

➡ cosec A

➡R.H.S.

That's Proved.

_________________

Answer 2

$\huge{\bold{\red{\mathtt{\underline{Solution}}}}}$

Given -

$\sec \alpha - \tan\alpha = x$

To prove -

$\frac{1 + {x}^{2} }{1 - {x}^{2} } = \csc \alpha$

Prove -

using these formulae -:

☆

${ \sec}^{2} \alpha - { \tan }^{2} \alpha = 1$

☆

$\tan \alpha = \frac{ \sin\alpha }{ \cos \alpha }$

☆

$\cos \alpha = \frac{1}{ \sec \alpha }$

LHS-:

$\frac{1 + {x}^{2} }{1 - {x}^{2} }$

keep the value of x in it -:

$\frac{1 + ( { \sec\alpha - \tan \alpha )}^{2} }{1 - ( { \sec\alpha - \tan \alpha )}^{2} }$

now putting the value of 1

$\frac{({ sec }^{2} \alpha - { \tan}^{2} \alpha)+ ( { \sec\alpha - \tan \alpha )}^{2} }{({ sec }^{2} \alpha - { \tan}^{2} \alpha) - ( { \sec\alpha - \tan \alpha )}^{2} }$

by opening the second bracket -:

$\frac{({ sec }^{2} \alpha - { \tan}^{2} \alpha)+ ( { { sec }^{2}\alpha + { \tan}^{2} - 2 sec\alpha \times tan\alpha )} }{({ sec }^{2} \alpha - { \tan}^{2} \alpha) - ( { {\sec}^2\alpha + {\tan }^2\alpha - 2 \tan\alpha \times \sec \alpha)} }$

by converting all the values of sec à into tan à -:

$\frac {2 ({sec}^2\alpha - sec\alpha \times tan\alpha)}{2 (-{tan}^2\alpha + sec\alpha \times tan\alpha)}$

by taking sec à and tan à common from then eq.

$\frac {sec\alpha (sec\alpha - tan\alpha)}{tan\alpha(-tan\alpha + sec\alpha)}$

as we can see sec à - tan à are common in the upper eq.

$\frac<u>{</u>sec\alpha}{tan\alpha}$

$sec \alpha$ = $\frac{1}{cos\alpha}$

$tan \alpha$ = $\frac{sin\alpha}{cos\alpha}$

putting these values -:

$\frac{\frac{1}{cos\alpha}}{\frac{sin\alpha}{cos\alpha}}$

$\frac{1}{sin\alpha}$

${csc\alpha}$

☆ hence proved ☆