Prove that: prove that 1/(secθ-tanθ)=secθ+tanθ
Chapter 6 – Trigonometry – Text Book Solution
Practice Set 6.1| Q 6.6 | Page 132
Prove that: prove that 1/(secθ-tanθ)=secθ+tanθ
Solution
We can start by using the definitions of secant and tangent:
$$frac{1}{sectheta – tantheta}$$
$$= frac{1}{frac{1}{costheta} – frac{sintheta}{costheta}}$$
$$= frac{costheta}{1 – sintheta}$$
Now, we can use the identity:
$$1 – sin^2theta = cos^2theta$$
To rewrite the denominator:
$$1 – sintheta = cos^2theta – sinthetacostheta$$
$$= costheta(costheta – sintheta)$$
So, we have:
$$frac{costheta}{costheta(costheta – sintheta)}$$
$$= frac{1}{costheta – sintheta}$$
$$= frac{costheta + sintheta}{costheta – sintheta} cdot frac{costheta + sintheta}{costheta + sintheta}$$
$$= frac{cos^2theta + 2sinthetacostheta + sin^2theta}{cos^2theta – sin^2theta}$$
$$= frac{1 + 2tanthetasectheta + tan^2theta}{sec^2theta – tan^2theta}$$
$$= frac{sectheta + tantheta)^2}{sec^2theta}$$
Therefore, we have shown that $$frac{1}{sectheta – tantheta} = sectheta + tantheta$$.
Solution
We can start by using the definitions of secant and tangent:
1/(secθ – tanθ)
= 1/(1/cosθ – sinθ/cosθ)
= cosθ/(1 – sinθ)
Now, we can use the identity:
1 – sin²θ = cos²θ
To rewrite the denominator:
1 – sinθ = cos²θ – sinθ*cosθ
= cosθ(cosθ – sinθ)
So, we have:
cosθ/(cosθ(cosθ – sinθ))
= 1/(cosθ – sinθ)
= (cosθ + sinθ)/(cosθ – sinθ) * (cosθ + sinθ)/(cosθ + sinθ)
= (cos²θ + 2sinθ cosθ + sin²θ)/(cos²θ – sin²θ)
= (1 + 2tanθ secθ + tan²θ)/(sec²θ – tan²θ)
= (secθ + tanθ)²/secθc²θ
Therefore, we have shown that 1/(secθ – tanθ) = secθ + tanθ.
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Chapter 6 – Trigonometry – Text Book Solution
Practice set 6.1 |Q 6.5| P 132
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