SSC CGL 3 Mar 2020 Shift 2 Paper

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Question : 52 of 75

Marks: +1, -0

If

\;{\ sec θ-\tan θ}/{\ sec θ+\tan θ}=\;{3}/{5}

, then the value of

\;{{\cosec} θ+\cot θ}/{{\cosec} θ-\cot θ}

is:

$27+{√{15}}$
$33+4 {√{ }} 15$
$31+8 {√{15}}$
$24+{√{15}}$

Solution:

Componendo and dividendo rule:
If

\;{ {\a}}/{ {\b}}=\;{ {\c}}/{ {\d}}

Then

\;{ {\a}+ {\b}}/{ {\a}- {\b}}=\;{ {\c}+ {\d}}/{ {\c}- {\d}}

(\ sec θ-\tan θ) /(\ sec θ+\tan θ)=3 / 5

⇒(\ sec θ+\tan θ) /(\ sec θ-\tan θ)=5 / 3

Using Componendo or Dividendo

{\ sec θ} / {\tan θ}=(5+3) /(5-3)

⇒(1 / {\cos θ}) /( {sin \;θ} / {\cos θ})=8 / 2

⇒ 1 / {sin \;θ}=4

⇒ sin \;θ=1 / 4

Perpendicular

= {\P}

, Hypotenuse

= {\H}

and Base

= {\B}

.

sin \;θ= {\P} / {\H}=1 / 4

{\P}=1

and

{\H}=4

As we know,

{\H}^2= {\P}^2+ {\B}^2

⇒ 4^2=1^2+ {\B}^2

⇒ {\B}^2=16-1=15

⇒ {\B}={√{ }} 15

{\cosec} θ=H / P=4

\cot θ=B / P={√{15}}

Now,

({\cosec} θ+\cot θ) \/({\cosec} θ-\cot θ)

(4+{√{ }} 15) /(4-{√{15}})

=\;{(4+{√{15}}) ×(4+{√{ }} 15)}/{(4-{√{15}}) ×(4+{√{1}} 5)}

=\;{(4+{√{15}})^2}/{(16-15)}

=(4+{√{ }} 15)^2

=16+15+2 × 4 {√{ }} 15

=3 1+8 {√{15}}

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