Prove that `(sinθ – cosθ + 1)/(sinθ + cosθ – 1) = 1/(secθ – tanθ)`
Chapter 6 – Trigonometry – Text Book Solution
Problem set 6| Q 5.10| Page 139
Prove that `(sinθ – cosθ + 1)/(sinθ + cosθ – 1) = 1/(secθ – tanθ)`
Solution
$$frac{sintheta-costheta+1}{sintheta+costheta-1}$$
$$=frac{(sintheta-costheta+1)(sintheta+costheta+1)}{(sintheta+costheta-1)(sintheta+costheta+1)}$$
$$=frac{(sintheta+1-costheta)(sintheta+1+costheta)}{(sin^2theta+cos^2theta+2sinthetacostheta-1)}$$
$$=frac{(sintheta+1)^2-cos^2theta}{(sin^2theta+cos^2theta+2sinthetacostheta-1)}$$
Since $$sin^2theta+cos^2theta=1$$, we get
$$frac{(1-cos^2theta+2sintheta)}{(2sinthetacostheta)}$$
$$=frac{(2-2cos^2theta+2sintheta)}{(2sinthetacostheta)}$$
$$=frac{(1-cos^2theta+sintheta)}{(sinthetacostheta)}$$
$$=frac{(sintheta+sin^2theta)}{(sinthetacostheta)}$$
$$=frac{(sintheta+1)}{costheta}$$
$$=frac{1}{costheta}+frac{sintheta}{costheta}$$
$$=sectheta+tantheta$$
$$=(sectheta+tantheta)frac{sectheta-tantheta}{sectheta-tantheta}$$
$$=frac{sec^2theta-tan^2theta}{sectheta-tantheta}$$
We know that$$sec^2theta-tan^2theta=1$$ and so we get
$$=frac{1}{sectheta-tantheta}$$
Hence, proved.
Solution
Consider the L.H.S.
(sinθ - cosθ + 1)/(sinθ + cosθ - 1)
= ((sinθ - cosθ + 1)/(sinθ + cosθ - 1)) * ((sinθ + cosθ + 1)/(sinθ + cosθ + 1))
= ((sinθ + 1 - cosθ)/(sinθ + cosθ - 1)) * ((sinθ + 1 + cosθ)/(sinθ + cosθ + 1))
= ((sinθ + 1)^2 - cos^2θ)/((sinθ + cosθ)^2 - 1^2)
= (sin^2θ + 1 + 2sinθ - cos^2θ)/(sin^2θ + cos^2θ + 2sinθcosθ - 1)
since sin^2θ + cos^2θ = 1
= (1 - cos^2θ + 1 + 2sinθ - cos^2θ)/(1 + 2sinθcosθ - 1)
= (2 - 2cos^2θ + 2sinθ)/(2sinθcosθ)
= (1 - cos^2θ + sinθ)/(sinθcosθ)
= (sin^2θ + sinθ)/(sinθcosθ)
= (sinθ + 1)/cosθ
= 1/cosθ + sinθ/cosθ
= secθ + tanθ
= (secθ + tanθ) * (secθ - tanθ)/(secθ - tanθ)
= (sec^2θ - tan^2θ)/(secθ - tanθ)
We know that sec^2θ – tan^2θ = 1
= 1/(secθ - tanθ)
Hence, it is proved.
Chapter 6 – Trigonometry – Text Book Solution
Problem Set 6 |Q 5.10| P 139
Click Here to see All the Textbook solution of Geometric Construction