1-sec2 -cos 2 - - 1-cosec2 -sin 2 - sin 2 - cos 2 - - 1-sin2 - cos 2 -2-sin 2 - cos2 -

{1/(sec^2 θ cos ^2 θ) + 1/(cosec^2 θ sin ^2 θ)} (sin ^2 θ cos ^2 θ) ={cos^2θ/(1+cos^2θ)sin^2θ + sin^2θ/(1+sin ^2 θ)cos^2θ} (sin ^2 θ cos ^2 θ) ={cos^4θ+cos^4θsi ...

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

Byju's Answer

Standard X

Mathematics

Trigonometric Identities

1/sec2 θ-cos ...

Question

${\frac{1}{(s e c^{2} θ - c o s^{2} θ)} + \frac{1}{(c o s e c^{2} θ - s i n^{2} θ)}} (s i n^{2} θ c o s^{2} θ) = \frac{1 - s i n^{2} θ c o s^{2} θ}{2 + s i n^{2} θ c o s^{2} θ}$

Open in App

Solution

${\frac{1}{(s e c^{2} θ - c o s^{2} θ)} + \frac{1}{(c o s e c^{2} θ - s i n^{2} θ)}} (s i n^{2} θ c o s^{2} θ)$

= ${\frac{{cos}^{2} θ}{(1 + {cos}^{2} θ) {sin}^{2} θ} + \frac{{sin}^{2} θ}{(1 + s i n^{2} θ) {cos}^{2} θ}} (s i n^{2} θ c o s^{2} θ)$

= ${\frac{{cos}^{4} θ + {cos}^{4} θ {sin}^{2} θ + {sin}^{4} θ + {sin}^{4} θ {cos}^{2} θ}{(1 + {cos}^{2} θ) (1 + {sin}^{2} θ)}}$

= ${\frac{{cos}^{4} θ + {sin}^{4} θ + {sin}^{2} θ {cos}^{2} θ ({sin}^{2} θ + {cos}^{2} θ)}{2 + {sin}^{2} θ {cos}^{2} θ}}$

= $\frac{1 - s i n^{2} θ c o s^{2} θ}{2 + s i n^{2} θ c o s^{2} θ}$

Suggest Corrections

139

Similar questions

\frac{sinθ + cosθ}{sinθ - cosθ} + \frac{sinθ - cosθ}{sinθ + cosθ} = \frac{2}{(\sin^{2} θ - \cos^{2} θ)} = \frac{2}{(2 \sin^{2} θ - 1)}

Q. If

y = {sin}^{2} θ + {cos}^{3} θ

. Find

\frac{d y}{d θ}

Q. Solve the following equations:
(i)

\cos θ + \cos 2 θ + \cos 3 θ = 0

(ii)

\cos θ + \cos 3 θ - \cos 2 θ = 0

(iii)

\sin θ + \sin 5 θ = \sin 3 θ

(iv)

\cos θ \cos 2 θ \cos 3 θ = \frac{1}{4}

(v)

\cos θ + \sin θ = \cos 2 θ + \sin 2 θ

(vi)

\sin θ + \sin 2 θ + \sin 3 = 0

(vii)

\sin θ + \sin 2 θ + \sin 3 θ + \sin 4 θ = 0

(viii)

\sin 3 θ - \sin θ = 4 \cos^{2} θ - 2

(ix)

\sin 2 θ - \sin 4 θ + \sin 6 θ = 0

Q. Simplify:-

\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}

Q. Prove the following trigonometric identities.

(\frac{1}{\sec^{2} θ - \cos^{2} θ} + \frac{1}{{cosec}^{2} θ - \sin^{2} θ}) \sin^{2} θ \cos^{2} θ = \frac{1 - \sin^{2} θ \cos^{2} θ}{2 + \sin^{2} θ \cos^{2} θ}

Join BYJU'S Learning Program

{1(sec2θ−cos2θ)+1(cosec2θ−sin2θ)}(sin2θcos2θ)=1−sin2θcos2θ2+sin2θcos2θ

{1(sec2θ−cos2θ)+1(cosec2θ−sin2θ)}(sin2θcos2θ) ={cos2θ(1+cos2θ)sin2θ+sin2θ(1+sin2θ)cos2θ}(sin2θcos2θ) ={cos4θ+cos4θsin2θ+sin4θ+sin4θcos2θ(1+cos2θ)(1+sin2θ)} ={cos4θ+sin4θ+sin2θcos2θ(sin2θ+cos2θ)2+sin2θcos2θ} =1−sin2θcos2θ2+sin2θcos2θ

${\frac{1}{(s e c^{2} θ - c o s^{2} θ)} + \frac{1}{(c o s e c^{2} θ - s i n^{2} θ)}} (s i n^{2} θ c o s^{2} θ) = \frac{1 - s i n^{2} θ c o s^{2} θ}{2 + s i n^{2} θ c o s^{2} θ}$