IS LHS-RHS- sin-cos-sin-cos-sin-cos-sin-cos- - 1-sin2- - cos2- - 2-tan2- - 1Say true or false-

The correct option is B FalseSimplifying, we get =(sinθ+cosθ)^2+(sinθ cosθ)^2sin^2θ cos^2θ =2sin^2θ cos^2θ =2cos^2θ(tan^2θ 1)=2sec^2θtan^2θ 1 ...

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Standard XII

Mathematics

Domain

IS LHS=RHS? ...

Question

IS LHS=RHS?

$\frac{sin θ + cos θ}{sin θ - cos θ} + \frac{sin θ - cos θ}{sin θ + cos θ} = \frac{1}{{sin}^{2} θ - {cos}^{2} θ} = \frac{{sec}^{2} θ}{{tan}^{2} θ - 1}$

Say true or false.

True

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False

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Ambiguous

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Data insufficient

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Solution

The correct option is B False
Simplifying, we get
$= \frac{(s i n θ + c o s θ)^{2} + (s i n θ - c o s θ)^{2}}{s i n^{2} θ - c o s^{2} θ}$
$= \frac{2}{s i n^{2} θ - c o s^{2} θ}$
$= \frac{2}{c o s^{2} θ (t a n^{2} θ - 1)} = \frac{2 s e c^{2} θ}{t a n^{2} θ - 1}$

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Similar questions

Q. Prove that

\frac{sin θ - cos θ}{sin θ + cos θ} + \frac{sin θ + cos θ}{sin θ - cos θ} = \frac{2}{{sin}^{2} θ - {cos}^{2} θ}

$Solve the following equations : (i) c o s θ + c p s 2 θ + c o s 3 θ = 0 (i i) c o s θ + c o s 3 θ - c o s 2 θ = 0 (i i i) s i n θ + s i n 5 θ = s i n 3 θ (i v) c o s θ c o s 2 θ c o s 3 θ = \frac{1}{4} (v) c o s θ + s i n θ = c o s 2 θ + s i n 2 θ (v i) s i n θ + s i n 2 θ + s i n 3 θ = 0 (v i i) s i n θ + s i n 2 θ + s i n 3 θ + s i n 4 θ = 0 (v i i i) s i n 3 θ - s i n θ = 4 c o s^{2} θ - 2 (i x) s i n 2 θ - s i n 4 θ + s i n 6 θ = 0$

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IS LHS=RHS?sinθ+cosθsinθ−cosθ+sinθ−cosθsinθ+cosθ=1sin2θ−cos2θ=sec2θtan2θ−1Say true or false.

The correct option is B FalseSimplifying, we get=(sinθ+cosθ)2+(sinθ−cosθ)2sin2θ−cos2θ=2sin2θ−cos2θ=2cos2θ(tan2θ−1)=2sec2θtan2θ−1

IS LHS=RHS?

$\frac{sin θ + cos θ}{sin θ - cos θ} + \frac{sin θ - cos θ}{sin θ + cos θ} = \frac{1}{{sin}^{2} θ - {cos}^{2} θ} = \frac{{sec}^{2} θ}{{tan}^{2} θ - 1}$

Say true or false.

The correct option is B False
Simplifying, we get
$= \frac{(s i n θ + c o s θ)^{2} + (s i n θ - c o s θ)^{2}}{s i n^{2} θ - c o s^{2} θ}$
$= \frac{2}{s i n^{2} θ - c o s^{2} θ}$
$= \frac{2}{c o s^{2} θ (t a n^{2} θ - 1)} = \frac{2 s e c^{2} θ}{t a n^{2} θ - 1}$