Hushar Mulga
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Prove That tanθ/(secθ – 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)

Chapter 6 – Trigonometry – Text Book Solution

Practice Set 6.1| Q 6.12 | Page 132

Prove That

tanθ/(secθ – 1) = (tanθ + secθ + 1)/(tanθ + secθ – 1)

Solution

We can start by simplifying the left-hand side of the equation:

tanθ/(secθ – 1)

= tanθ/(1/cosθ – 1) (using the identity secθ = 1/cosθ)

= tanθ/((1 – cosθ)/cosθ)

= (sinθ/cosθ) / ((1 – cosθ)/cosθ)

= sinθ/(cosθ – cos²θ)

= sinθ/(sin²θ/cosθ)

= cosθ/sinθ

= cotθ

Next, we can simplify the right-hand side of the equation:

(tanθ + secθ + 1)/(tanθ + secθ – 1)

= ((sinθ/cosθ) + (1/cosθ) + 1)/((sinθ/cosθ) + (1/cosθ) – 1)

= ((sinθ + 1 + cosθ)/cosθ)/((sinθ + 1)/cosθ)

= (sinθ + cosθ + 1)/(sinθ + 1)

Now, we can see if the two sides are equal by cross-multiplying:

cotθ = (sinθ + cosθ + 1)/(sinθ + 1)

(sinθ + cosθ)/sinθ = (sinθ + cosθ + 1)/sinθ

sinθ + cosθ = sinθ + cosθ + 1

1 = 1

Therefore, we have shown that the two sides are equal and proved that:

tanθ/(secθ – 1) = (tanθ + secθ + 1)/(tanθ + secθ – 1)

Chapter 6 – Trigonometry – Text Book Solution

Practice set 6.1 |Q 6.12| P 132

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