Question

verify that the equation is an identity.
\\( \sin ^ { 2 } \alpha \sec ^ { 2 } \alpha + \sin ^ { 2 } \alpha \csc ^ { 2 } \alpha = \sec ^ { 2 } \alpha \\)
to verify the identity, start with the more complicated side and transform it to look like the other side. choose the correct transformations and
\\( \sin ^ { 2 } \alpha \sec ^ { 2 } \alpha + \sin ^ { 2 } \alpha \csc ^ { 2 } \alpha \\)
\\( = \sin ^ { 2 } \alpha ( \square ) \\)
rewrite in terms of the cosine function.
separate a single - term into two terms.
factor out the greatest common factor.
rewrite in terms of the sine function.

Explanation:

Step1: Factor out the greatest common factor

We have \(\sin^{2}\alpha\sec^{2}\alpha+\sin^{2}\alpha\csc^{2}\alpha\). By the distributive property \(ab + ac=a(b + c)\), where \(a = \sin^{2}\alpha\), \(b=\sec^{2}\alpha\) and \(c=\csc^{2}\alpha\), we can factor out \(\sin^{2}\alpha\). So \(\sin^{2}\alpha\sec^{2}\alpha+\sin^{2}\alpha\csc^{2}\alpha=\sin^{2}\alpha(\sec^{2}\alpha+\csc^{2}\alpha)\)

Step2: Rewrite in terms of sine and cosine functions

We know that \(\sec\alpha=\frac{1}{\cos\alpha}\) and \(\csc\alpha=\frac{1}{\sin\alpha}\). Then \(\sec^{2}\alpha+\csc^{2}\alpha=\frac{1}{\cos^{2}\alpha}+\frac{1}{\sin^{2}\alpha}\)

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Since \(\sin^{2}\alpha+\cos^{2}\alpha = 1\), we have \(\frac{\sin^{2}\alpha+\cos^{2}\alpha}{\sin^{2}\alpha\cos^{2}\alpha}=\frac{1}{\sin^{2}\alpha\cos^{2}\alpha}\)

Step3: Simplify the expression

We have \(\sin^{2}\alpha(\sec^{2}\alpha+\csc^{2}\alpha)=\sin^{2}\alpha\times\frac{1}{\sin^{2}\alpha\cos^{2}\alpha}\)
Cancel out \(\sin^{2}\alpha\) (assuming \(\sin\alpha
eq0\)), we get \(\frac{1}{\cos^{2}\alpha}=\sec^{2}\alpha\)

Answer:

The given equation \(\sin^{2}\alpha\sec^{2}\alpha+\sin^{2}\alpha\csc^{2}\alpha=\sec^{2}\alpha\) is an identity.