Class 10 exercise 8.4 ques 5

(i) (cosec θ – cot θ)² = (1-cos θ)/(1+cos θ)

To prove this, first take the Left-Hand side (L.H.S) of the given equation, to prove the Right Hand Side (R.H.S)

L.H.S. = (cosec θ – cot θ)²

The above equation is in the form of (a-b)², and expand it

Since (a-b)² = a² + b² – 2ab

Here a = cosec θ and b = cot θ

= (cosec²θ + cot²θ – 2cosec θ cot θ)

Now, apply the corresponding inverse functions and equivalent ratios to simplify

= (1/sin²θ + cos²θ/sin²θ – 2cos θ/sin²θ)

= (1 + cos²θ – 2cos θ)/(1 – cos²θ)

= (1-cos θ)²/(1 – cosθ)(1+cos θ)

= (1-cos θ)/(1+cos θ) = R.H.S.

Therefore, (cosec θ – cot θ)² = (1-cos θ)/(1+cos θ)

Hence proved.

(ii)  (cos A/(1+sin A)) + ((1+sin A)/cos A) = 2 sec A

Now, take the L.H.S of the given equation.

L.H.S. = (cos A/(1+sin A)) + ((1+sin A)/cos A)

            = [cos²A + (1+sin A)²]/(1+sin A)cos A

            = (cos²A + sin²A + 1 + 2sin A)/(1+sin A) cos A

Since cos²A + sin²A = 1, we can write it as

            = (1 + 1 + 2sin A)/(1+sin A) cos A

            = (2+ 2sin A)/(1+sin A)cos A

            = 2(1+sin A)/(1+sin A)cos A

            = 2/cos A = 2 sec A = R.H.S.

L.H.S. = R.H.S.

(cos A/(1+sin A)) + ((1+sin A)/cos A) = 2 sec A

Hence proved.

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<a href="https://mycbseguide.com/blog/ncert-solutions-class-10-maths-exercise-8-4/" ping="/url?sa=t&source=web&rct=j&url=https://mycbseguide.com/blog/ncert-solutions-class-10-maths-exercise-8-4/&ved=2ahUKEwiAh6LGlLPsAhWhQ3wKHRY_BY0QFjADegQIAhAC" rel="noopener" target="_blank">NCERT Solutions for Class 10 Maths Exercise 8.4 ...</a>