Question

it is for std 10 lesson no 6 practice 6.1 Q . no 6 (4) sum if u known the answer then only reply answer

Answer 1

GIVEN:

(sec θ - cos θ)(cot θ + tan θ) = tan θ sec θ

TO PROVE:

(sec θ - cos θ)(cot θ + tan θ) = tan θ sec θ

PROOF:

$\green{\boxed{ \large{ \bold{sec \: \theta = \frac{1}{cos \: \theta}}}}}$

$\left( \dfrac{1}{cos \: \theta} - cos \: \theta\right)cot \: \theta +tan \: \theta$

$\left( \dfrac{1 - {cos }^{2} \theta}{cos \: \theta}\right)cot \: \theta +tan \: \theta$

$\red{\boxed{ \large{ \bold{1 - {cos }^{2} \theta = {sin}^{2} \: \theta}}}}$

$\left( \dfrac{{sin}^{2} \theta}{cos \: \theta}\right)cot \: \theta +tan \: \theta$

$\blue{\boxed{ \large{ \bold{cot \: \theta = \frac{ cos \: \theta}{sin \: \theta}}}}}$

$\orange{\boxed{ \large{ \bold{tan \: \theta = \frac{ sin \: \theta}{cos \: \theta}}}}}$

$\left( \dfrac{{sin}^{2} \theta}{cos \: \theta}\right)\left( \dfrac{{sin} \theta}{cos \: \theta} + \dfrac{cos \: \theta}{sin \: \theta}\right)$

$\left( \dfrac{{sin}^{2} \theta}{cos \: \theta}\right)\left( \dfrac{{sin ^{2} } \theta + {cos}^{2} \theta }{cos \: \theta \: sin \: \theta} \right)$

$\purple{\boxed{ \large{ \bold{{{sin ^{2} } \theta + {cos}^{2} \theta = 1}}}}}$

$\left( \dfrac{sin\theta}{cos \: \theta}\right)\left( \dfrac{1}{cos \: \theta} \right)$

$tan \: \theta \: \sec \: \theta$

L.H.S = R.H.S

(sec θ - cos θ)(cot θ + tan θ) = tan θ sec θ

HENCE PROVED.

Answer 2

Answer:

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