Question

Q: Prove It :-


( tan 2ⁿ Φ )/tan Φ = (1 + sec 2Φ)(1 + sec 2²Φ)(1 + sec 2³Φ)(1 + sec 2⁴Φ)... (1 + sec 2ⁿΦ)​

Answer 1

Answer:

tan2 θ – (1/cos2 θ) + 1 = 0

Solution:

L.H.S = tan2 θ – (1/cos2 θ) + 1

= tan2 θ - sec2 θ + 1 [since, 1/cos θ = sec θ]

= tan2 θ – (1 + tan2 θ) +1 [since, sec2 θ = 1 + tan2 θ]

= tan2 θ – 1 – tan2 θ + 1

= 0 = R.H.S. Proved

2. Verify that:

1/(sin θ + cos θ) + 1/(sin θ - cos θ) = 2 sin θ/(1 – 2 cos2 θ)

Solution:

L.H.S = 1/(sin θ + cos θ) + 1/(sin θ - cos θ)

= [(sin θ - cos θ) + (sin θ + cos θ)]/(sin θ + cos θ)(sin θ - cos θ)

= [sin θ - cos θ + sin θ + cos θ]/(sin2 θ - cos2 θ)

= 2 sin θ/[(1 - cos2 θ) - cos2 θ] [since, sin2 θ = 1 - cos2 θ]

= 2 sin θ/[1 - cos2 θ - cos2 θ]

= 2 sin θ/[1 – 2 cos2 θ] = R.H.S. Proved

3. Prove that:

sec2 θ + csc2 θ = sec2 θ ∙ csc2 θ

Solution:

L.H.S. = sec2 θ + csc2 θ

= 1/cos2 θ + 1/sin2 θ [since, sec θ = 1/cos θ and csc θ = 1/sin θ]

= (sin2 θ + cos2 θ)/(cos2 θ sin2 θ)

= 1/cos2 θ ∙ sin2 θ [since, sin2 θ + cos2 θ = 1]

= 1/cos2 θ ∙ 1/sin2 θ

= sec2 θ ∙ csc2 θ = R.H.S. Proved

Step-by-step explanation:

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Answer 2

$\large\underline{\sf{Solution-}}$

Let assume that

$\sf \:P(n):(1 + sec2\phi )(1 + {sec2}^{2}\phi ) - - - (1 + {sec2}^{n}\phi) = \dfrac{tan {2}^{n} \phi }{tan\phi }$

We use Principal of Mathematical Induction to prove this.

Step :- 1 For n = 1

$\rm :\longmapsto\:\sf \:P(1):(1 + sec2\phi )$

$\rm \:  =  \: 1 + \dfrac{1}{cos2\phi }$

$\rm \:  =  \: \dfrac{cos2\phi + 1}{cos2\phi }$

$\rm \:  =  \: \dfrac{1 + \dfrac{1 - {tan}^{2}\phi }{1 + {tan}^{2} \phi } }{\dfrac{1 - {tan}^{2}\phi }{1 + {tan}^{2} \phi }}$

$\rm \:  =  \: \dfrac{\dfrac{1 + {tan}^{2}\phi + 1 - {tan}^{2}\phi }{1 + {tan}^{2} \phi } }{\dfrac{1 - {tan}^{2}\phi }{1 + {tan}^{2} \phi }}$

$\rm \:  =  \: \dfrac{2}{1 - {tan}^{2}\phi }$

$\rm \:  =  \: \dfrac{2tan\phi }{1 - {tan}^{2}\phi } \times \dfrac{1}{tan\phi }$

$\rm \:  =  \: \dfrac{tan2\phi }{tan\phi }$

$\bf\implies \:P(n) \: is \: true \: for \: n = 1$

Step :- 2 Assume that P(n) is true for n = k where k is natural number

$\sf \:P(k):(1 + sec2\phi )(1 + {sec2}^{2}\phi ) - - - (1 + {sec2}^{k}\phi) = \dfrac{tan {2}^{k} \phi }{tan\phi }$

Step :- 3 Now, we have to prove that P(n) is true for n = k + 1

$\sf \:P(k + 1):(1 + sec2\phi )(1 + {sec2}^{2}\phi ) - - - (1 + {sec2}^{k + 1}\phi) = \dfrac{tan {2}^{k + 1} \phi }{tan\phi }$

Consider, LHS

$\sf \:(1 + sec2\phi )(1 + {sec2}^{2}\phi ) - - - (1 + {sec2}^{k + 1}\phi)$

$\sf \: = \: \dfrac{tan {2}^{k} \phi }{tan\phi }(1 + {sec2}^{k + 1}\phi)$

$\sf \: = \: \dfrac{tan {2}^{k} \phi }{tan\phi }\bigg(1 + \dfrac{1}{cos {2}^{k + 1} \phi } \bigg)$

$\sf \: = \: \dfrac{tan {2}^{k} \phi }{tan\phi }\bigg( \dfrac{1 + {cos2}^{k + 1} \phi }{cos {2}^{k + 1} \phi } \bigg)$

$\sf \: = \: \dfrac{tan{2}^{k} \phi }{tan\phi }\bigg(\dfrac{1 + \dfrac{1 - {tan}^{2}{2}^{k}\phi }{1 + {tan}^{2}{2}^{k} \phi }}{\dfrac{1 - {tan}^{2}{2}^{k}\phi }{1 + {tan}^{2} {2}^{k}\phi }} \bigg)$

$\sf \: = \: \dfrac{tan {2}^{k} \phi }{tan\phi }\bigg(\dfrac{\dfrac{1 + {tan2}^{k}\phi + 1 - {tan}^{2k}\phi }{1 + {tan}^{2k} \phi }}{\dfrac{1 - {tan}^{2k}\phi }{1 + {tan}^{2k} \phi }} \bigg)$

$\sf \: = \: \dfrac{tan {2}^{k} \phi }{tan\phi }\bigg(\dfrac{\dfrac{2}{1 + {tan}^{2}{2}^{k} \phi }}{\dfrac{1 - {tan}^{2}{2}^{k}\phi }{1 + {tan}^{2} {2}^{k}\phi }} \bigg)$

$\sf \:  =  \: \dfrac{tan {2}^{k} \phi }{tan\phi } \times \dfrac{2}{1 - {tan}^{2}{2}^{k} \phi }$

$\sf \:  =  \: \dfrac{1 }{tan\phi } \times \dfrac{2 {tan2}^{k} \phi }{1 - {tan}^{2}{2}^{k} \phi }$

$\sf \:  =  \: \dfrac{tan2.( {2}^{k} \phi )}{tan\phi }$

$\sf \:  =  \: \dfrac{tan {2}^{k + 1} \phi }{tan\phi }$

$\bf\implies \:P(n) \: is \: true \: for \: n = k + 1$

Hence, By the Process of Principal of Mathematical Induction,

$\sf \:(1 + sec2\phi )(1 + {sec2}^{2}\phi ) - - - (1 + {sec2}^{n}\phi) = \dfrac{tan {2}^{n} \phi }{tan\phi }$

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Formula Used

$\boxed{\tt{ cos2\phi = \dfrac{1 - {tan}^{2}\phi }{1 + {tan}^{2} \phi }}}$

$\boxed{\tt{ tan2\phi = \dfrac{2tan\phi }{1 - {tan}^{2} \phi }}}$