Question

Prove that, (sin a + cos a) (tan a + cot a) = sec a + cosec a
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Answer 1

Given

Prove that (sinA + cosA) (tanA + cotA) = secA + cosecA

To Prove

(sinA + cosA) (tanA + cotA) = secA + cosecA

$\rule{200}2$

Proof

Taking L.H.S.

⇒ (sinA + cosA) (tanA + cotA)

We know that, tanA = sinA/cosA and cotA = cosA/sinA

⇒ (sinA + cosA) (sinA/cosA + cosA/sinA)

⇒ (sinA + cosA) [(sin²A + cos²A)/cosA.sinA]

Also, sin²A + cos²A = 1

⇒ (sinA + cosA) (1/cosA.sinA)

⇒ (sinA + cosA)/(cosA.sinA)

⇒ sinA/cosA.sinA + cosA/cosA.sinA

⇒ 1/cosA + 1/sinA

We know that, 1/cosA = secA and 1/sinA = cosecA

⇒ secA + cosecA

L.H.S. = R.H.S.

Hence, proved.

Answer 2

❥${\red{\underline{\huge{\mathtt{Question:-}}}}}$

Prove that, (sin a + cos a) (tan a + cot a) = sec a + cosec a.

❥${\red{\underline{\huge{\mathtt{Solution:-}}}}}$

${\purple{\bold{\huge{L.H.S}}}}$

${\bold{(sin a + cos a)\: (tan a + cot a)}}$

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★Write tan a = sin a/cos a

cot a = cos a/ sin a★

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${\bold{→(sin a + cos a)\: (\frac{sin\:a}{cos\:a}+\frac{cos\:a}{sin\:a})}}$

${\bold{→(sin a + cos a)\: (\frac{sin^2a+cos^2a}{cos\:a.sin\:a})}}$

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★ We know sin²a + cos²a = 1★

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${\bold{→(sin a + cos a)\: \frac{1}{cos\:a.sin\:a}}}$

${\bold{→\frac{sin\:a+cos:a}{cos\:a.sin\:a}}}$

${\bold{→\frac{sin\:a}{cos\:a.sin\:a}+\frac{cos\:a}{cos\:a.sin\:a}}}$

${\bold{→\frac{1}{cos\:a}+\frac{1}{sin\:a}}}$

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★We know 1/ cos a= sec a and 1/sin a= cosec a ★

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${\green{\bold{→sec\:a+cosec\:a}}}$

L.H.S = R.H.S

Hence proved______!!