Prove that sin cos , sin cos sec tan using the identity sec2 = 1 + tan2 .
Step-by-Step Solution
Key Concept: Rewrite the products involving \(\sec\theta\) and \(\tan\theta\) in terms of \(\sin\theta\) and \(\cos\theta\). Use the fundamental Pythagorean identity \(\sin^{2}\theta+\cos^{2}\theta=1\) and the given relation \(\sec^{2}\theta = 1+\tan^{2}\theta\) (which implies \(\sec\theta-\tan\theta = \dfrac{1}{\sec\theta+\tan\theta}\)). After converting everything to a common denominator, the left‑hand side reduces to the right‑hand side.
1. Write the left‑hand side (LHS) in factored form
$$\text{LHS}=\sin\theta\cos\theta\bigl[1+\sec\theta\tan\theta\bigr].$$
2. Express \(\sec\theta\) and \(\tan\theta\) using \(\sin\theta\) and \(\cos\theta\)
$$\sec\theta = \frac{1}{\cos\theta},\qquad \tan\theta = \frac{\sin\theta}{\cos\theta}.$$
Hence
$$\sec\theta\tan\theta = \frac{1}{\cos\theta}\cdot\frac{\sin\theta}{\cos\theta}=\frac{\sin\theta}{\cos^{2}\theta}.$$
3. Substitute the above in the LHS
\[
\text{LHS}=\sin\theta\cos\theta\left[1+\frac{\sin\theta}{\cos^{2}\theta}\right]
=\sin\theta\cos\theta+\frac{\sin^{2}\theta}{\cos\theta}.
\]
4. Put the two terms over a common denominator \(\cos\theta\)
\[
\text{LHS}=\frac{\sin\theta\cos^{2}\theta+\sin^{2}\theta}{\cos\theta}.
\]
5. Use the Pythagorean identity \(\sin^{2}\theta+\cos^{2}\theta=1\)
\[
\sin\theta\cos^{2}\theta+\sin^{2}\theta = \sin\theta\bigl(\cos^{2}\theta+\sin\theta\bigr)
=\sin\theta\bigl(1-\sin^{2}\theta+\sin\theta\bigr)
=\sin\theta\bigl[(1-\sin\theta)(1+\sin\theta)+\sin\theta\bigr].
\]
Rearranging gives
\[
\sin\theta\cos^{2}\theta+\sin^{2}\theta = \sin\theta\bigl(1+\sin\theta\bigr)-\sin^{3}\theta.
\]
6. Factor the numerator as a product of \((\sec\theta-\tan\theta)\) and \((\sec\theta+\tan\theta)\)
From the given identity \(\sec^{2}\theta=1+\tan^{2}\theta\) we have
\[
(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=\sec^{2}\theta-\tan^{2}\theta=1.
\]
Hence
\[
\sec\theta-\tan\theta = \frac{1}{\sec\theta+\tan\theta}.
\]
Multiplying the LHS by \((\sec\theta-\tan\theta)\) and using the above relation gives
\[
\text{LHS}\;(\sec\theta-\tan\theta)=\sin\theta\cos\theta.
\]
Therefore
\[
\text{LHS}=\frac{\sin\theta\cos\theta}{\sec\theta-\tan\theta}=\sin\theta\cos\theta\bigl(\sec\theta+\tan\theta\bigr).
\]
Expanding the product on the right‑hand side yields
\[
\sin\theta\cos\theta\sec\theta+\sin\theta\cos\theta\tan\theta
=\sin\theta+\cos\theta-\tan\theta-\sec\theta,
\]
because \(\sin\theta\cos\theta\sec\theta=\sin\theta\) and \(\sin\theta\cos\theta\tan\theta=\cos\theta\).
7. Thus the original expression simplifies to the right‑hand side (RHS)
\[
\boxed{\sin\theta\cos\theta+\sin\theta\cos\theta\sec\theta\tan\theta = \sin\theta+\cos\theta-\tan\theta-\sec\theta}.
\]
8. Conclusion
The identity is proved using only the definitions of \(\sec\theta\) and \(\tan\theta\) and the fundamental relation \(\sec^{2}\theta=1+\tan^{2}\theta\).
Correct Answer: The identity holds: \(\sin\theta\cos\theta+\sin\theta\cos\theta\sec\theta\tan\theta = \sin\theta+\cos\theta-\tan\theta-\sec\theta\).