Question

tanA/secA-1+tanA/secA+1=2cosecA(trigonometry Identities)​

Answer 1

Answer:



simm4923

15.01.2018

Math

Secondary School

+5 pts

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Prove that tanA/secA-1+tanA/secA+1=2cosecA

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Ambitious

LHS = tanA/secA - 1 + tanA/secA + 1

= sinA/cosA / 1/cosA - 1 + sinA/cosA / 1/cosA + 1 [because tan A = sin A / cos A and sec A = 1 / cos A]

= sinA/cosA / (1 - cosA) / cosA + sinA/cosA / (1 + cosA) / cos A

= sinA / 1 - cosA + sinA / 1 + cos A

= [sinA (1 + cosA) + sinA (1 - cosA)] / (1 - cosA) (1 + cosA)

= [sinA + sinAcosA + sinA - sinAcosA] / 1 - cos(sq) A

= 2sinA / sin(sq) A [From identity: sin(sq) A + cos(sq) A = 1]

= 2 / sin A

= 2 x 1 / sinA

=2 cosecA

Answer 2

Correct QuestioN :

$\tt \dagger \: \: \: \: \: \dfrac{tan\, A}{1 + Sec\, A} - \dfrac{ tan\,A}{1 - Sec\,A} = 2Cosec\,A.$

SolutioN :

☯ Taking LHS.

$\tt \dagger \: \: \: \: \: \dfrac{tan\, A}{1 + Sec\, A} - \dfrac{ tan\,A}{1 - Sec\,A}$

$\tt : \implies \dfrac{tan\, A}{1 + Sec\, A} - \dfrac{ tan\,A}{1 - Sec\,A}$

$\tt : \implies tan \,A\Bigg\lgroup\dfrac{1}{1 + Sec\, A} - \dfrac{1 }{1 - Sec\,A} \Bigg\rgroup$

$\tt : \implies tan \,A\Bigg\lgroup\dfrac{(1 - Sec\, A) - (1 + Sec\, A)}{(1 + Sec\, A)( 1 - Sec\,A)} \Bigg\rgroup$

$\tt : \implies tan \,A\Bigg\lgroup\dfrac{1 - Sec\, A- 1 - Sec\, A}{(1 + Sec\, A)( 1 - Sec\,A)} \Bigg\rgroup$

$\tt : \implies tan \,A\Bigg\lgroup\dfrac{ - Sec\, A- Sec\, A}{(1 + Sec\, A)( 1 - Sec\,A)} \Bigg\rgroup$

$\tt : \implies tan \,A\Bigg\lgroup\dfrac{- 2 Sec\, A}{(1 + Sec\, A)( 1 - Sec\,A)} \Bigg\rgroup$

$\tt : \implies \Bigg\lgroup\dfrac{- 2 Sec\, A \times tan \,A}{ {(1)}^{2} - {Sec\, }^{2} A}\Bigg\rgroup$

♢ We can taking negative sign common.

$\tt : \implies \Bigg\lgroup\dfrac{- 2 Sec\, A \times tan \,A}{ - ( {Sec\, }^{2} A - 1)}\Bigg\rgroup$

$\tt : \implies \Bigg\lgroup\dfrac{2 Sec\, A \times tan \,A}{ {Sec\, }^{2} A - 1}\Bigg\rgroup$

☛ Now, We Know that.

• 1 + tan²A = sec²A.

$\tt : \implies \Bigg\lgroup\dfrac{2 Sec\, A \times tan \,A}{ {tan\, }^{2} A }\Bigg\rgroup$

$\tt : \implies \Bigg\lgroup\dfrac{2 Sec\, A \times \cancel{ tan \,A}}{ \cancel{ {tan\, }^{2} A}}\Bigg\rgroup$

$\tt : \implies \dfrac{2 Sec\, A}{ tan\, A }$

☛ We can also write as,

• Sec A = 1 / Cos A.
• tan A = Sin A / Cos A.

$\tt : \implies \dfrac{2 \times \frac{1}{cos\, A \: } }{ \frac{sin\, A}{ coa\, A } }$

$\tt : \implies \dfrac{2}{Cos\,A} \times \dfrac{Cos\,A}{Sin\, A}$

$\tt : \implies \dfrac{2}{Sin\, A}$

$\tt : \implies 2Cosec\,A.$

✡ Hence Proved.