RRB NTPC Mains or Stage 2 Model Paper 5 with solutions

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Question : 40 of 120

Marks: +1, -0

The value of

(1 + sec 20° + cot 70°) (1 – cosec 20° + tan 70°)

is

–1
2
1
0

Solution:

(b)

(1 + sec 20° + cot 70°) (1 – cosec 20° + tan 70°)

= [1 + cosec (90° – 20°) + cot 70°]

[1 – sec (90° – 70°) + tan 70°]

= [1 + cosec 70° + cot 70°] [1 – sec 70° + tan 70°]

=[1 +1/\text" sin 70°"+\text" cos70 °"/\text"sin 70°"]

[1-1/\text"cos70°"+\text"sin 70°" /\text"cos70°"]

=

\text"(sin^270°+ cos70°+1)( sin70°+ cos70°-1)"/\text"sin70° cos70°"

=\text"(sin^2 70°+ cos^270° +2sin70° cos70 °-1)"/\text"sin70° cos70°"

=\text"2sin70° cos70°"/\text"sin 70° cos70°"

[∴π sin^2 70° + cos^2 70° = 1]= 2

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