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Prove that (sin(90^circ – θ))/(cos θ) + (cos(90^circ – θ))/(sin θ) = 2. - Mathematics

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Question

Prove that `(sin(90^circ - θ))/(cos θ) + (cos(90^circ - θ))/(sin θ) = 2`.

Theorem
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Solution

Given: Let θ be an angle. Consider `sin(90^circ - θ)/(cos θ) + (cos (90^circ - θ))/(sin θ)`

To Prove: `sin(90^circ - θ)/(cos θ) + (cos (90^circ - θ))/(sin θ) = 2`

Proof [Step-wise]:

1. Use the complementary-angle identities:

sin (90° – θ) = cos θ and cos (90° – θ) = sin θ

2. Substitute these into the expression:

`sin(90^circ - θ)/(cos θ)`

= `(cos θ)/(cos θ)`

= 1

3. Similarly, `cos(90^circ - θ)/(sin θ)`

= `(sin θ)/(sin θ)`

= 1

4. Add the two results:

1 + 1 = 2

Hence, `sin(90^circ - θ)/(cos θ) + (cos (90^circ - θ))/(sin θ) = 2`, as required.

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Chapter 18: Trigonometric Ratios of Some Standard Angles and Complementary Angles - Exercise 18C [Page 380]

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Nootan Mathematics [English] Class 9 ICSE
Chapter 18 Trigonometric Ratios of Some Standard Angles and Complementary Angles
Exercise 18C | Q 16. (i) | Page 380
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