tan θ + tan (90° - θ) = sec θ sec (90° - θ). Prove the following statement
Solution:
We know that
tan (90° - θ) = cot θ
LHS = tan θ + tan (90° - θ)
= tan θ + cot θ
We know that
tan θ = sin θ/ cos θ
cot θ = cos θ/ sin θ
LHS = sin θ/ cos θ + cos θ/ sin θ
Taking LCM
= (sin2 θ + cos2 θ)/ sin θ cos θ
As sin2 θ + cos2 θ = 1
= 1/ sin θ cos θ
It can be written as
= 1/ cos θ x 1/sin θ
= sec θ cosec θ
Here
cosec θ = sec (90° - θ)
= sec θ sec (90° - θ)
= RHS
Therefore, it is proved.
✦ Try This: Simplify: (cos(90° + θ) sec(-θ) tan(180° - θ))/(sec(360° - θ) sin(180° + θ) cot(90° + θ))
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 8
NCERT Exemplar Class 10 Maths Exercise 8.3 Problem 7
tan θ + tan (90° - θ) = sec θ sec (90° - θ). Prove the following statement
Summary:
The secant function of a right triangle is its hypotenuse divided by its base. It is proved that tan θ + tan (90° - θ) = sec θ sec (90° - θ)
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