The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is
a. -1
b. 0
c. 1
d. 3/2

Solution:

We have to find the value of the expression [cosec (75° + θ) - sec (15° - θ) - tan (55° + θ) + cot (35° - θ)]

We know that cosec A = sec (90° - A)

Also, cot A = tan (90° - A)

Now, cosec (75° + θ) = sec (90° - (75° + θ))

= sec (15° - θ)

cot (35° - θ) = tan (90° - (35° - θ))

= tan (55°+ θ)

So, [cosec (75° + θ) - sec (15° - θ) - tan (55° + θ) + cot (35° - θ)]

= [sec (15° + θ) - sec (15° - θ) - tan (55° + θ) + tan (55° + θ)]

= 0

Therefore, the value of the expression [cosec (75° + θ) - sec (15° - θ) - tan (55° + θ) + cot (35° - θ)] is zero.

✦ Try This: Prove that tan (55° - θ) - cot (35° + θ) = 0

LHS = tan (55° - θ) - cot (35° + θ)

We can write

cot A = tan (90° - A)

= tan (90° - (35° + θ)) - cot (35° + θ)

= cot (35° + θ) - cot (35° + θ)

= 0

= RHS

Therefore, it is proved.

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 8

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NCERT Exemplar Class 10 Maths Exercise 8.1 Problem 3

The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is a. -1, b. 0, c. 1, d. 3/2

Summary:

The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is zero

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☛ Related Questions:

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• If sin A = 1/2 , then the value of cot A is a. √3, b. 1/√3, c. √3/2, d. 1