Question

Tan ^3 theta / 1+tan ^2 theta + cot^3 theta /1+cot ^2 theta = sec theta × cosec theta - 2sin theta × cos theta

Answer 1

Answer:

Step-by-step explanation:

Tan*3/1+tan*2 + cot*3/1+cot*2

Sin*3/cos*3÷1+sin*2+cos² + cos³/sin³÷1+cos²/sin²

Sin³/cos³÷ sin²+cos²/cos² + cos³/sin³ ÷ sin² + cos²

Sin³/cos³ ÷ 1/cos² + cos³/sin³ ÷ 1/sin²

Sin³/cos. + cos³/sin

Sin*4+ cos*4 ÷ sin × cos

(Sin²)² + (cos²)² ÷ sin × cos

(Sin²+ cos²)² - 2 sin²×cos² ÷ sin×cos

1-2sin²×cos²÷ sin×cos

1/sin×cos -2sin²×cos²/sin×cos

Sec×cosec - 2sin×cos

Lhs=Rhs

Hence proved

Hope it helps

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Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: QUESTION \: \: } \mid}}}}}$

$\tt{ \: \: Prove \: \: \: that, \: \: } \\ \\ \tt{\frac{tan {}^{3 } \theta }{1 + tan {}^{2} \theta} + \frac{cot {}^{3 } \theta }{1 + cot{}^{2} \theta} = sec \: \theta \: cosec \: \theta- 2 \: sin \: \theta \: \cos \: \theta}$

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\tt{L.H.S. = \frac{tan {}^{3 } \theta }{1 + tan {}^{2} \theta} + \frac{cot {}^{3 } \theta }{1 + cot{}^{2} \theta} } \\ \\ \tt{ = \frac{ \frac{sin {}^{3} \theta}{cos {}^{3} \theta} }{sec {}^{2} \theta} + \frac{ \frac{cos {}^{3} \theta}{sin {}^{3} \theta} }{cosec {}^{2} \theta}} \\ \\ \tt{ = ( \frac{sin {}^{3} \theta}{cos {}^{3} \theta } \times \frac{1}{sec {}^{2} \theta } ) + ( \frac{cos {}^{3} \theta}{sin {}^{3} \theta } \times \frac{1}{cosec {}^{2} \theta } ) } \tt{ = ( \frac{sin {}^{3} \theta}{cos {}^{3} \theta } \times cos {}^{2} \theta) + ( \frac{cos {}^{3} \theta}{sin {}^{3} \theta } \times sin {}^{2} \theta ) }$

$\tt{ = \frac{sin {}^{3} \theta}{cos \: \theta} + \frac{cos {}^{3} \theta}{sin \: \theta} } \\ \\ \tt{ = \frac{sin {}^{4} \theta + cos {}^{4} \theta}{sin \: \theta .cos \: \theta} }\\ \\ \tt{= \frac{sin {}^{4} \theta + cos {}^{4} \theta + 2.sin {}^{2} \theta.cos {}^{2} \theta -2.sin {}^{2} \theta.cos {}^{2} \theta }{sin \: \theta .cos \: \theta} } \tt{ = \frac{(sin {}^{2} \theta + cos {}^{2} \theta) {}^{2} - 2.sin {}^{2} \theta.cos {}^{2} \theta}{sin \: \theta . cos \: \theta}}$

$\tt{= \frac{(1) {}^{2} - 2.sin {}^{2} \theta.cos {}^{2} \theta }{sin \: \theta. cos \: \theta}} \\ \\ \tt{ = \frac{1 - 2.sin {}^{2} \theta.cos {}^{2} \theta }{sin \: \theta .cos \: \theta}} \\ \\ \tt{ = \frac{1}{sin \: \theta.cos \: \theta} - \frac{2.sin {}^{2} \theta .cos {}^{2} \theta}{sin \: \theta .cos \: \theta} } \\ \\ \tt{= \frac{1}{sin \: \theta} . \frac{1}{cos \: \theta }- 2 \: sin \: \theta \: \cos \: \theta }\\ \\ \tt{ = cosec \: \theta \:sec \: \theta- 2 \: sin \: \theta \: \cos \: \theta } \tt{= sec \: \theta \: cosec \: \theta- 2 \: sin \: \theta \: \cos \: \theta} \\ \\ \tt{ = R.H.S.}$

FORMULA USED

$\tt{1. \: \: tan \: A = \frac{sin \: A}{cos \: A} \: \implies \: tan {}^{2} \: A = \frac{sin {}^{2} \: A}{cos {}^{2} \: A} } \\ \\ \tt{2. \: \: cot \: A = \frac{cos \: A}{sin\: A} \: \implies \: cot{}^{2} \: A = \frac{cos {}^{2} \: A}{sin {}^{2} \: A} } \\ \\ \tt{3. \: \:1 + tan {}^{2} A = sec {}^{2} A } \\ \\ \tt{4. \: \:1 + cot{}^{2} A = cosec {}^{2} A } \\ \\ \tt{5. \: \: \frac{1}{sec \: A} = cos \: A } \\ \\ \tt{6. \: \: \frac{1}{cosec \:A } = sin \: A }$