Sin -cot -cosec - - sin -cot -cosec -2

Sin θ/(cot θ+cosec θ) sin θ/(cot θ cosec θ)=2 LHS =sin θ/(cot θ+cosec θ) sin θ/(cot θ cosec θ) = sin θ(cot θ cosec θ) sin θ(cot θ+cosec θ)/(cot θ+cosec ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Trigonometric Identities

sin θ/cot θ+c...

Question

$\frac{s i n θ}{(c o t θ + c o s e c θ)} - \frac{s i n θ}{(c o t θ - c o s e c θ)} = 2$

Open in App

Solution

$\frac{s i n θ}{(c o t θ + c o s e c θ)} - \frac{s i n θ}{(c o t θ - c o s e c θ)} = 2 L H S = \frac{s i n θ}{(c o t θ + c o s e c θ)} - \frac{s i n θ}{(c o t θ - c o s e c θ)} = \frac{s i n θ (c o t θ - c o s e c θ) - s i n θ (c o t θ + c o s e c θ)}{(c o t θ + c o s e c θ) (c o t θ - c o s e c θ)} = \frac{s i n θ c o t θ - s i n θ c o s e c θ - s i n θ c o t θ - s i n θ c o s e c θ}{c o t^{2} θ - c o s e c^{2} θ} = \frac{- 2 s i n θ c o s e c θ}{- 1} = 2 \times s i n θ \times \frac{1}{s i n θ} = 2 = R H S$

Suggest Corrections

Similar questions

Q. Solve :

\frac{1}{cosec θ + cot θ} - \frac{1}{sin θ} = \frac{1}{sin θ} - \frac{1}{cosec θ - cot θ}

Q. Show that

\frac{1}{cosec θ - cot θ} - \frac{1}{sin θ} = \frac{1}{sin θ} - \frac{1}{cosec θ + cot θ}

Q. If

\frac{cosec θ + cot θ}{cosec θ - cot θ} = \frac{81}{49}

, then find

\frac{cos θ + sin θ}{sin θ - cos θ}

Q. Is LHS=RHS?

\frac{1}{c o s e c θ + cot θ} - \frac{1}{sin θ} = \frac{1}{sin θ} - \frac{1}{c o s e c θ - cot θ}

Q. If sin

θ

\frac{\sqrt{2}}{3}

, cos

θ

\frac{1}{3}

Find tan

θ

, cot

θ

, sec

θ

and cosec

θ

Join BYJU'S Learning Program

sinθ(cotθ+cosecθ)−sinθ(cotθ−cosecθ)=2

sinθ(cotθ+cosecθ)−sinθ(cotθ−cosecθ)=2LHS=sinθ(cotθ+cosecθ)−sinθ(cotθ−cosecθ)=sinθ(cotθ−cosecθ)−sinθ(cotθ+cosecθ)(cotθ+cosecθ)(cotθ−cosecθ)=sinθ cotθ−sinθ cosecθ−sinθ cotθ−sinθ cosecθcot2θ−cosec2θ=−2sinθ cosecθ−1=2×sinθ×1sinθ=2=RHS

$\frac{s i n θ}{(c o t θ + c o s e c θ)} - \frac{s i n θ}{(c o t θ - c o s e c θ)} = 2$