Prove 3 - 4sin2-cos2- - 3 - tan 2-

LHS = 3 4sin^2θ #160; cos^2θ #10; #10; 3 ^2θ #160; 4tan ^2θ #10; #10; 3(1 + tan ^2θ ) 4tan #10;^2θ #10; #10; 3 + 3ta ...

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Trigonometric Identities

Prove 3 - 4...

Question

Prove $\frac{3 - 4 {sin}^{2} θ}{{cos}^{2} θ} = 3 - {tan}^{2} θ$

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Solution

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\frac{3 - 4 {sin}^{2} Θ}{{cos}^{2} Θ}

\frac{3 - 4 {sin}^{2} θ}{{cos}^{2} θ}

\frac{s i n 2 θ}{1 - c o s 2 θ} = c o t θ o r \frac{1 - c o s^{2} θ}{1 + c o s 2 θ} = \frac{t a n^{2} θ}{2}

If $sec θ = \frac{17}{8}$ , verify that $\frac{3 - 4 s i n^{2} θ}{4 c o s^{2} θ - 3} = \frac{(3 - t a n^{2} θ)}{(1 - 3 t a n^{2} θ)}$ .

4 {sin}^{2} θ - 1 = 0

90^{\circ}

{cos}^{2} θ + {tan}^{2} θ

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