A Prove that cos A-1-sin A-tan A- Ab Prove the identity sin-cos-tan-cosec- -6 MARKS-

Each Part: 3 Marks (a) cos A/1+sin A + tan A = sec A L.H.S = cos A/1+sin A + tan A =cos A/1+sin A×1 sin A/1 sin A + tan A =c̸o̸s̸ ̸A̸(̸1̸ ̸s̸i̸n̸ ̸A̸)̸/cos^ ...

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Byju's Answer

Standard X

Mathematics

Relation between Trigonometric Ratios

a Prove that ...

Question

(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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Solution

Each Part: 3 Marks

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

L.H.S $= \frac{c o s A}{1 + s i n A} + t a n A$

$= \frac{c o s A}{1 + s i n A} \times \frac{1 - s i n A}{1 - s i n A} + t a n A$

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

$= \frac{1 - / s i n A + / s i n A}{c o s A}$

$= \frac{1}{c o s A} = s e c A = R . H . S$

(b) LHS $= (s i n θ + c o s θ) (t a n θ + c o t θ)$

$= (s i n θ + c o s θ) (\frac{s i n θ}{c o s θ} + \frac{c o s θ}{s i n θ})$

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

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

$= \frac{s i n θ + c o s θ}{c o s θ s i n θ} = \frac{s i n θ}{c o s θ s i n θ} + \frac{c o s θ}{c o s θ s i n θ}$

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

$= s e c θ + c o s e c θ =$ RHS

LHS = RHS

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Similar questions

Q. Prove that :

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

Prove the identity $[cosec (90^{\circ} - θ) - sin (90^{\circ} - θ)] [cosec θ - sin θ] [tan θ + cot θ] = 1$

Prove the following trigonometric identities.
(i) $\frac{1 + cos θ + sin θ}{1 + cos θ - sin θ} = \frac{1 + sin θ}{cos θ}$

(ii) $\frac{sin θ - cos θ + 1}{sin θ + cos θ - 1} = \frac{1}{sec θ - tan θ}$

(iii) $\frac{cos θ - sin θ + 1}{cos θ + sin θ - 1} = cosec θ + cot θ$

(iv) $(sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ$

Q. (i) Prove that

\frac{c o t θ + c o s e c θ - 1}{c o t θ - c o s e c θ + 1} = c o s e c θ + c o t θ = \frac{1 + c o s θ}{s i n θ}

(ii) Prove that

\frac{1}{s e c A - t a n A} - \frac{1}{c o s A} = \frac{1}{c o s A} - \frac{1}{s e c A + t a n A}

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

Join BYJU'S Learning Program

(a) Prove that cos A1+sin A+tan A=sec A (b) Prove the identity (sin θ+cos θ)(tan θ+cot θ)=sec θ+cosec θ [6 MARKS]

(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]