Question

write the following functions in the simplest form :
1•tan-¹(cosx - sin x/cos x + sin x)
2• tan-¹(3a²x-x³/a³-3ax²)
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Answer 1

Question :-

• tan-¹(cosx - sin x/cos x + sin x)

Solution :-

→ tan-¹(cosx - sin x/cos x + sin x)

Divide Both Numerator & Denominator by cosx we get ,

→ tan-¹[{(cosx/cosx) - (sin x/cosx)} /{(cosx/cosx) + (sin x/cosx)}]

Using (sinx/cosx) = tanx now,

→ tan-¹[( 1 - tanx ) / (1 + tanx)]

Or,

→ tan-¹[ (1 - tanx) / (1 + 1 * tanx) ]

using 1 = tan(π/4) now,

→ tan-¹[ ( tan(π/4) - tanx) / (1 + tan(π/4)*tanx) ]

Comparing it with [ (tanA - tanB) / ( 1 + tanA*tanB) ]

→ tan-¹[ tan( π/4 - x ) ]

→ (π/4 - x) (Ans.)

Answer 2

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$\huge\tt{TO~SIMPLIFY:}$

• tan-¹(cosx - sin x/cos x + sin x)
• tan-¹(3a²x-x³/a³-3ax²)

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$\huge\tt{SIMPLIFYING:}$

1. tan-¹(cosx - sin x/cos x + sin x)

➠tan -¹ (cosx - sinx/cosx + sinx)

➠tan -¹{(cosx/cosx)-(sinx/cosx)/(cosx/cosx) + (sinx/cosx}

➠tan-¹ [(1-tanx)/(1+tanx)]

➠tan-¹[(1-tanx)/(1+1×tanx)]

➠tan -¹[{tan(π/4)-tanx}/{1+tan(π/4)×tanx}]

➠tan -¹ [tan(π/4-x)]

➠(π/4-x)

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2. tan-¹(3a²x-x³/a³-3ax²)

➠tan-¹[{3a²(a tan∅)-(a tan∅)³}/{a³-3 tan ²∅}]

➠tan-¹[{3a³ tan∅ - a³ tan³ ∅}/{a³- 3a.a²tan²∅}]

➠tan-¹[{a³ (3tan∅ - tan³ ∅}/{a³- (1-3tan²∅)}]

➠tan-¹[{3tan∅-tan³∅}/{1-3tan²∅}]

➠tan-¹(tan3∅)

➠3∅

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