If \tan 15^\circ = 2 -\sqrt 3 , the value of tan 15°. cot 75° + tan 75° . cot 15° is.

RRB NTPC Mains or Stage 2 Model Paper 7 with solutions

Question : 20 of 120

Marks: +1, -0

tan 15° = 2 -√ 3 ,

the value of tan 15°. cot 75° + tan 75° . cot 15° is.

Solution:

(a) tan 15° . cot 75° + tan 75° . cot 15°
= tan 15 . cot (90° – 15°) + tan (90° – 15°) cot 15°
= tan 15° . tan 15° + cot 15° . cot 15°

= (tan 15)^2 + (cot 15)^2

= (tan 15°)^2 +1/( tan)^2

Putting the value of

tan 15° = 2 - √3

=(2 -√3)^2 +[1/2 -√3]^2

=(2- √3)^2 +[1/ 2- √3×2+√3/2+√3]^2

= (2 - √3)^2 + (2 + √3/4-3)^2

=(2-√3)^2+(2+√3)^2

[∵ (a + b)^2 + (a – b)^2 = 2 (a^2 + b^2)]

= 2 ( 4 + 3) = 2 × 7 = 14

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