Question

#QualityQuestion
@Trigonometry


$\tt{Prove ~the~ identity ~:}$
tan²(x) - sin²(x) = tan²(x) sin²(x)​

Answer 1

~Solution :-

$\large \mathfrak{LHS :}$

$\sf{ta {n}^{2} (x) - si {n}^{2} (x) = si {n}^{2} (x) / co {s}^{2} (x) - si {n}^{2} (x)}$

$\to \sf{ \frac{[si {n}^{2} (x) - co {s}^{2} (x) si {n}^{2} (x) ] } {co {s}^{2} (x)}}$

$\to \sf{ \frac{ si {n}^{2} (x) [ 1 - co {s}^{2} (x) ] }{ co {s}^{2} (x)}}$

$\to \sf{ \frac{si {n}^{2} (x) si {n}^{2} (x)}{ co {s}^{2} (x)}}$

$\implies \bf \red{ si {n}^{2} (x) ta {n}^{2} (x)}$

$\large \mathfrak{ : RHS}$

— BrainlyROME —

Answer 2

Assuming tan²(x) - sin²(x) = tan²(x)sin²(x),

start off by rewriting tan²(x) into it's sin(x) and cos (x) components.

$\frac{{\sin}^{2}(x)}{{\cos}^{2}(x)} - {\sin}^{2}(x)$

Finding a common denominator

$\frac{{\sin}^{2}(x)}{{\cos}^{2}(x)}\times\frac{1}{1} - {\sin}^{2}(x)\times\frac{{\cos}^{2}(x)}{{\cos}^{2}(x)}\rightarrow\frac{{\sin}^{2}(x)}{{\cos}^{2}(x)} - \frac{{\sin}^{2}(x){\cos}^{2}(x)}{{\cos}^{2}(x)}$

Combine in to a single fraction and factor out a ${\sin}^{2}(x)$.

$\frac{{\sin}^{2}(x) - {\sin}^{2}(x){\cos}^{2}(x)}{{\cos}^{2}(x)}\rightarrow {\sin}^{2}(x)\times\frac{{\sin}^{2}(x)}{{\cos}^{2}(x)}$

Finally write

${\sin}^{2}(x)\times\frac{{\sin}^{2}(x)}{{\cos}^{2}(x)}\rightarrow{\sin}^{2}(x){\tan}^{2}(x)$

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