Question

show that 1 - sin²48°/ cos²42° ( sec²42° - 1) = tan²45°
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Answer 1

✨✨✨ Thanks for asking ✨✨

ANSWER :➡➡➡➡

Step by step explanation :

$rhs \\ \frac{1 - { \sin }^{2} 48}{ { \cos}^{2} 42( { \sec}^{2} 42 - 1)} \\ = \frac{ { \cos}^{2}48 }{ \cos^{2} 42 \times { \tan }^{2} 42} ( \sin^{2} \alpha + \cos ^{2} \alpha = 1) \\ \: \: \: so \: ( \cos^{2} \alpha = 1 - { \sin }^{2} \alpha )\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \: \: \: \: \: ( \sec^{2} \alpha - 1 = { \tan}^{2} \alpha ) \\ \\ = \frac{ { \cos}^{2} (90 - 42)}{ \cos^{2} 42 \times { \tan}^{2} 42} = \frac{ { \sin }^{2} 42}{ { \cos }^{2} 42} \times \frac{1}{ { \tan }^{2} 42} ↩↩\\ = {tan}^{2} 42\times \frac{1}{ {tan}^{2} 42} \: \: \: \: \: ( \frac{ {sin}^{2} }{ {cos}^{2} } = tan^2)\\ = 1 \\ lhs \\ we \: know \: that \\ \: \: \: \: \: \: \: \: \: \: {tan}^{2} 45 = {1}^{2} = 1\\ so \: \: \: lhs = \: rhs \\ \\ hence \: \frac{1 - {sin}^{2} 48}{ {cos}^{2}42( {sec}^{2} 42 - 1)} = {tan}^{2} 45$

✨✨✨hope it helps✨✨✨