Question

solve this plz......................... ​

Answer 1

$\huge\mathcal{\underline{\red{\underline{\blue{ Solution:-}}}}}$

${tan}^{2}\theta - \frac{1}{{cos}^{2}\theta} + 1 = 0\\ \\</p><p>LHS = {tan}^{2}\theta - \frac{1}{{cos}^{2}\theta} + 1 \\ \\ </p><p> = \frac{{sin}^{2}\theta}{{cos}^{2}\theta} - \frac{1}{{cos}^{2}\theta} + 1 \\ \\ </p><p> = \frac{{sin}^{2}\theta - 1}{{cos}^{2}} + 1 \\ \\ </p><p> = \frac{{sin}^{2}\theta - 1 + {cos}^{2}\theta}{{cos}^{2}} \\ \\</p><p> \boxed{\red{\boxed{\pink{{sin}^{2} + {cos}^{2}= 1 }}}} \\ \\ </p><p> = \frac{1 - 1}{{cos}^{2}} \\ \\</p><p> = \frac{0}{{cos}^{2}} \\ \\ = 0\: RHS \\ \\ \bold{HENCE \;PROVED}$

Answer 2

Answer:

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