RRB NTPC 29 April 2016 Paper

Question : 62 of 100

Marks: +1, -0

If 4sinθ - 3cosθ = 0, then secθ cosecθ is :

Solution:

Here, 4 sinθ = 3cosθ
So, 4 sinθ - 3cosθ = 0

\tan θ = { 3 } { 4 } = { P } /{ B }

H ^ { 2 } = 3 ^ { 2 } + 4 ^ { 2 } = 5 ^ { 2 }

H = 5

\sec θ. \cosec θ = { H }/ { B } × { H } /{ P } = { 5 } /{ 3 } × { 5 }/ { 4 } = { 25 }/ { 12 }

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