Prove that:(secθ – cosθ)(cotθ + tanθ) = tanθ.secθ.
Chapter 6 – Trigonometry – Text Book Solution
Practice Set 6.1| Q 6.4 | Page 132
Prove that:(secθ – cosθ)(cotθ + tanθ) = tanθ.secθ.
Solution
To prove: $$(sectheta – costheta)(cottheta + tantheta) = tantheta sectheta$$
We know that $$cottheta = frac{costheta}{sintheta}$$ and $$tantheta = frac{sintheta}{costheta}$$. Substituting these in the given equation, we get:
$$(sectheta – costheta)left(frac{costheta}{sintheta} + frac{sintheta}{costheta}right) = frac{sintheta}{costheta} cdot frac{1}{costheta}$$
Multiplying both sides by $$frac{sintheta costheta}{costheta}$$, we get:
$$(sectheta – costheta)(cos^2theta + sin^2theta) = sintheta$$
Using the identity $$cos^2theta + sin^2theta = 1$$, we get:
$$(sectheta – costheta) = frac{sintheta}{1} = sintheta$$
Multiplying both sides by $$frac{costheta}{sintheta}$$, we get:
$$(sectheta – costheta)(cottheta + tantheta) = tantheta sectheta$$
Therefore, we have shown that $$(sectheta – costheta)(cottheta + tantheta) = tantheta sectheta$$.
Solution
We can start by expanding the left-hand side of the equation using the definitions of secant, cosine, cotangent, and tangent:
(secθ – cosθ)(cotθ + tanθ)
= secθcotθ + secθtanθ – cosθcotθ – cosθtanθ
= (1/cosθ)(cosθ/sinθ) + (1/cosθ)(sinθ/cosθ) – cosθ*(cosθ/sinθ) – cosθ*(sinθ/cosθ)
= (1/sinθ) + (sinθ/cos²θ) – cos²θ/sinθ – sinθ
= [(1/cos²θ) – cos²θ]/sinθ + sinθ/cos²θ
= (1 – cos⁴θ)/sinθcos²θ + sinθ/cos²θ
= [(1 – cos²θ)(1 + cos²θ)]/sinθcos²θ
= (sin²θ/sinθ)(1/cos²θ)
= sinθ/cosθ
= tanθ.secθ
Therefore, we have shown that (secθ – cosθ)(cotθ + tanθ) = tanθ.secθ.
Explanation:-
Chapter 6 – Trigonometry – Text Book Solution
Practice set 6.1 |Q 6.4 | P 132
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