Question

※ HEY BUDDIES ※ I am with a trigonometry question here : Prove that ( Sin A + Cosec A ) ² + ( Cos A + Sec A ) ² = ( 7 Tan ² A + Cot ² A ) Please solve it friends , I need it for my exam !!

Thanks in advance ☺
@Beautiful68 !!

Answer 1

$\bold{\large{\underline{Hey\: mate \:! }}}$

$\bold{\large{\underline{Solution\: \colon}}}$

$\bold{\large{\underline{Revise \: formulae \: \colon}}}$

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$\bold{\large{1) \: cos^{2} \: \theta + sin^{2}\: \theta \: = \: 1}}$

$\bold{\large{2) \: cosec\: \theta = \frac{1}{sin\: \theta}}}$

$\bold{\large{3) \: sec \: \theta = \frac{1}{cos\: \theta}}}$

$\bold{\large{4) \: 1 + cot^{2}\: \theta = cosec^{2}\: \theta}}$

$\bold{\large{ 5) \: 1 + tan^{2} \: \theta = tan^{2}\: \theta }}$

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$\bold{\large{Q) \: (sin\: A +cosec\: A )^{2} + (cos\: A + sec\: A )^{2} = 7 + tan^{2} A + cot^{2} A}}$

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$\bold{\underline{Take\: LHS \: \colon}}$

Use ( a + b )² = a² + b² +2 ab

$\implies \bold{\large{( sin^{2}\: A + cosec^{2} \: A + 2 sin \: A . cosec \: A ) + ( cos^{2}\: A + sec^{2} \: A + 2 cos \: A . Sec \: A )}}$

Use formula 2 & 3

$\implies \bold{\large{[ sin^{2}\: A + cos^{2} \: A ] + [ cosec^{2}\: A + sec^{2}\: A] +[2\frac{sin\: A }{sin\: A } + 2\frac{cos\: A }{cos\: A }]}}$

After applying formula (1)

$\implies \bold{\large{ [1] \: + \: 2 +\: 2 + [ cosec^{2} A \: + \: sec^{2} \: A ]}}$

Apply formula (4) & (5)

$\implies \bold{\large{[5] \: + \: [ ( 1 \: + \: cot^{2} \: A ) + ( 1\: + tan^{2} \: A ) ]}}$

$\implies \bold{\large{ [ 5 \: +\: 1 \: + \: 1 ] + tan^{2} \: A\: + \: cot^{2} \: A }}$

$\implies \bold{\large{ [ 7 ] \: + \: tan^{2} \: A \: + \: cot^{2} \: A }}$

$\implies \bold{\large{7 \: + \: tan^{2} \: A \: + \: cot^{2} \: A }}$

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$\bold{\large{\boxed{ LHS \: = \: RHS }}}$

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Error in your RHS please correct it

$\bold{\large{ Thanks}}$

Answer 2

Answer:

how do you verify answers??