Question

Prove The Following.

i )
$\frac{tan³∅ - 1}{tan∅ - 1 } = sec²∅ + tan∅$

ii )
$\frac{sin∅-cos∅+1}{sin∅-cos∅ - 1} = \frac{1}{sec∅ \: - tan \: ∅}$
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Answer 1

Answer:

hard

hardStep-by-step explanation:

i )

$\frac{tan³∅ - 1}{tan∅ - 1 } = sec²∅ + tan∅$

ii )

$\frac{sin∅-cos∅+1}{sin∅-cos∅ - 1} = \frac{1}{sec∅ \: - tan \: ∅}$

Answer 2

Question:-

Prove The Following.

$\sf \: i ) \sf \frac{tan³∅ - 1}{tan∅ - 1 } \sf= sec²∅ + tan∅$

$\sf \: ii ) \sf\frac{sin∅-cos∅+1}{sin∅-cos∅ - 1} \sf= \frac{1}{sec∅ \: - tan \: ∅}$

Given:-

$\sf \: i ) \sf \frac{tan³∅ - 1}{tan∅ - 1 } \sf= sec²∅ + tan∅$

$\sf \: ii ) \sf\frac{sin∅-cos∅+1}{sin∅-cos∅ - 1} \sf= \frac{1}{sec∅ \: - tan \: ∅}$

To Find:-

Needed to prove the Following Equation.

Question:-

i)

$\sf LHS = \frac{tan {}^{3} \theta - 1 }{tan \theta - 1} \sf = \frac{(tan \theta - 1)(tan {}^{2} \theta + tan \theta + 1) }{(tan \theta - 1)}$

= tan²θ + tanθ + 1

[∵a³ – b³ = (a – b) (a² + ab + b²) ]

= (1 + tan²θ) + tanθ = sec²θ + tanθ = RHS.

Answer:-

$\sf \fbox\red{ =(1 + tan²θ) + tanθ = sec²θ + tanθ = RHS. }$

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ii)

$\sf LHS = \frac{sin \theta - cos \theta + 1}{sin \theta + cos \theta - 1} = \frac{tan \theta - 1 + sec \theta}{tan \theta + 1 - sec \theta}$

$\sf = \frac{(tan {}^{2} \theta - sec {}^{2} \theta) - (tan \theta - sec \theta) }{ (tan \theta - sec \theta + 1)(tan \theta - sec \theta)}$

$\sf = \frac{ - 1 - tan \theta + sec \theta}{(tan \theta - sec \theta + 1)(tan \theta - sec \theta)}$

$\sf = \frac{ { \cancel{-}} 1}{ { \cancel{-}} (sec \theta - tan \theta)} = \frac{1}{sec \theta - tan \theta} \: \: RHS$

Answer:-

$\sf {= \frac{ { \cancel{-}} 1}{ { \cancel{-}} (sec \theta - tan \theta)} = \frac{1}{sec \theta - tan \theta} \: \: RHS}$

$\sf \huge \fbox \red{Hence, Proved.}$

Hope you have satisfied. ⚘