RRB NTPC Mains or Stage 2 Model Paper 8 with solutions

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Question : 40 of 120

Marks: +1, -0

If

tan θ= -1/√ 7

, then

{cosec^2θ- sec^2θ}/ {cosec^2 θ+sec^2θ } = ?

$-3/ 4$
$– 2 /3$
$2/ 3$
$3/ 4$

Solution:

(d) tan θ =

{BC}/{AB}= 1 /√7=1/ {√7k}

AC^2 = (AB^2 + BC^2) = (7k^2 + k^2) = 8k^2

∴AC = √{8k^2 }= 2 √2k

sec θ

= {AC}/{AB} ={2√ 2k}/K= 2√ 2

,
cosec θ

= {AB} /{BC}={2 √2k} /K=2√ 2

∴{cosec^2θ- sec ^2θ}/ {cosec ^2θ+sec ^2 θ}={(2 √2)^ 2-[{{2√2}/ √7}]^2}/{(2 √2)^ 2+[{{2√2}/ √7}]^2}

=[ {8- 8/7 }/8+8/ 7] = 48/64=3/4

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