A Show that -1-cos A -1-cos A -sin A -1-cos Ab Prove that tan 2-12-1-cos-1-cos-6 MARKS-

Each Part: 3 Marks (a) LHS = √(1 cos A/1+cos A) =√((1 cos A)(1+cos A)/(1+cos A)(1+cos A)) =√(1 cos^2 A/(1+cos A)^2) = √(sin^2 A/(1+cos A)^2) =sin A/1+cos A ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

a Show that √...

Question

(a) Show that $\sqrt{\frac{1 - c o s A}{1 + c o s A}} = \frac{s i n A}{1 + c o s A}$

(b) Prove that $\frac{t a n^{2} θ}{(s e c θ - 1)^{2}} = \frac{1 + c o s θ}{1 - c o s θ}$ [6 MARKS]

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Solution

Each Part: 3 Marks

(a) LHS $= \sqrt{\frac{1 - c o s A}{1 + c o s A}}$

$= \sqrt{\frac{(1 - c o s A) (1 + c o s A)}{(1 + c o s A) (1 + c o s A)}}$

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

$= \frac{s i n A}{1 + c o s A} =$ RHS Proved

(b) LHS $= \frac{t a n^{2} θ}{(s e c θ - 1)^{2}} = {[\frac{\frac{s i n θ}{c o s θ}}{\frac{1}{c o s θ}} - 1]}^{2}$

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

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

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

$= \frac{1 + c o s θ}{1 - c o s θ} =$ RHS

Hence proved

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Similar questions

Q. Prove that

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

Q. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i)

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

(ii)

\frac{cos A}{1 + sin A} + \frac{1 + sin A}{cos A} = 2 sec A

(iii)

\frac{tan θ}{1 - cot θ} + \frac{cot θ}{1 - tan θ} = 1 + sec θ cosec θ

[Hint: Write the expression in terms of

sin θ

and

cos θ

(iv)

\frac{1 + sec A}{sec A} = \frac{{sin}^{2} A}{1 - cos A}

[Hint: Simplify LHS and RHS separately].

(v)

\frac{cos A - sin A - 1}{cos A + sin A + 1} = cosec A + cot A

, Using the identity

{cosec}^{2} A = 1 + {cot}^{2} A

(vi)

\sqrt{\frac{1 + sin A}{1 - sin A}} = sec A + tan A

(vii)

\frac{sin θ - 2 {sin}^{3} θ}{2 {cos}^{3} θ - cos θ}

(viii)

{(sin A + cosec A)}^{2} + {(cos A + sec A)}^{2} = 7 + {tan}^{2} A + {cot}^{2} A

(ix)

(cosec A - sin A) (sec A - cos A) = \frac{1}{tan A + cot A}

[Hint: Simplify LHS and RHS separately]

(x)

(\frac{1 + {tan}^{2} A}{1 + {cot}^{2} A}) = {(\frac{1 - tan A}{1 - cot A})}^{2} = {tan}^{2} A

Q. Prove that :

\frac{1 + cos A + sin A}{1 + cos A - sin A} = \frac{1 + sinA}{cos A}

Q. Prove that

\frac{sin A + 1 - cos A}{sin A - 1 + cos A} = \frac{1 + sin A}{cos A}

Q. Prove the following identities , where the angles involved are acute angles for which the expressions are defined .
(i)

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

(ii)

\frac{c o s A}{1 + s i n A} + \frac{1 + s i n A}{c o s A} = 2 s e c A

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(a) Show that √1−cos A1+cos A=sin A1+cos A (b) Prove that tan2 θ(sec θ−1)2=1+cos θ1−cos θ [6 MARKS]

(a) Show that $\sqrt{\frac{1 - c o s A}{1 + c o s A}} = \frac{s i n A}{1 + c o s A}$

(b) Prove that $\frac{t a n^{2} θ}{(s e c θ - 1)^{2}} = \frac{1 + c o s θ}{1 - c o s θ}$ [6 MARKS]