Prove that- - -sin- -cos-tan- -1

Prove that (cosec θ sinθ )(θ cosθ )(tanθ + θ ) = 1 LHS =(cosecθ #160; sinθ )(θ cosθ )(tanθ + θ ) = ( 1/sinθ sinθ #160; ) ( 1/cosθ ...

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Standard X

Mathematics

Trigonometric Ratios of a Right Triangle

Prove that: ...

Question

Prove that:
$(csc θ - sin θ) (sec θ - cos θ) (tan θ + cot θ) = 1$

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Solution

Prove that
$(c o s e c θ - sin θ) (sec θ - cos θ) (tan θ + cot θ) = 1$
LHS $= (c o s e c θ - sin θ) (sec θ - cos θ) (tan θ + cot θ)$
$= (\frac{1}{sin θ} - sin θ) (\frac{1}{cos θ} - cos θ)$ $(\frac{sin θ}{cos θ} + \frac{cos θ}{sin θ})$
$= (\frac{1 - {sin}^{2} θ}{sin θ}) (\frac{1 - {cos}^{2} θ}{cos θ}) (\frac{{sin}^{2} θ + {cos}^{2} θ}{sin θ . cos θ})$
$= \frac{{cos}^{2} θ}{sin θ} . \frac{{sin}^{2} θ}{cos θ} (\frac{1}{sin θ . cos θ})$
$= sin θ . cos θ . (\frac{1}{sin θ . cos θ})$
$= 1$
= RHS

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Similar questions

Q. Prove the following identities (1-17)

(cosec θ - \sin θ) (\sec θ - \cos θ) (\tan θ + \cot θ) = 1

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

(cosec θ - sin θ) (sec θ - cos θ) (tan θ + cot θ) = . . . . . . .

Q. Prove that :

(sin θ + cos θ) (tan θ + cot θ) = sec θ + csc θ

(a) Prove that $\frac{c o s A}{1 + s i n A} + t a n A = s e c A$

(b) Prove the identity $(s i n θ + c o s θ) (t a n θ + c o t θ) = s e c θ + c o s e c θ$ [6 MARKS]

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Prove that:(cscθ−sinθ)(secθ−cosθ)(tanθ+cotθ)=1

Prove that(cosecθ−sinθ)(secθ−cosθ)(tanθ+cotθ)=1LHS=(cosecθ−sinθ)(secθ−cosθ)(tanθ+cotθ)=(1sinθ−sinθ)(1cosθ−cosθ)(sinθcosθ+cosθsinθ)=(1−sin2θsinθ)(1−cos2θcosθ)(sin2θ+cos2θsinθ.cosθ)=cos2θsinθ.sin2θcosθ(1sinθ.cosθ)=sinθ.cosθ.(1sinθ.cosθ)=1= RHS

Prove that:
$(csc θ - sin θ) (sec θ - cos θ) (tan θ + cot θ) = 1$