Question

1+tan²A upon 1+secA = secA
prove
trignometry identities
pls solve step by step ​

Answer 1

Solution:

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Here in this question use appropriate trigonometry identities and solve it. In this case we will use trigonometry identities [ tan² A = sec² A - 1 ]. In this type of problems we will modify L.H.S to R.H.S which has to be proved.

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L.H.S

$\sf = 1 + \dfrac{ {\tan}^{2} A}{1 + \sec A }$

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• Use trigonometry Identity

➛ tan² A = sec² A - 1

$\\ \sf = 1 + \dfrac{ {({\sec}^{2} A - 1) }}{1 + \sec A }$

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• Use identity

➛ a² - b² = ( a + b ) ( a - b )

$\\ \sf = 1 + \dfrac{ {(\sec A - 1) (\sec A + 1)}}{1 + \sec A }$

$\\ \sf = 1 + \dfrac{ { (\sec A - 1) \cancel{(\sec A + 1)}}}{ \cancel{1 + \sec A}}$

$\\ \sf = 1 + \sin A - 1$

$\\ \sf = \sec A$

= R.H.S

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L.H.S = R.HS

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HENCE PROVED!!

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Some more trigonometry identities:

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• sin² θ + cos² θ = 1
• sin² θ = 1 - cos² θ

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• 1 + tan² θ = sec² θ
• tan² θ = sec² θ - 1

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• sec² A - tan² A = 1

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• cosec² A - cot² A = 1

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