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Values of Trigonometric Ratios

Question 5 v ...

Question

Question 5 (v)
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(v) $\frac{(c o s A - s i n A + 1)}{(c o s A + s i n A - 1)} = c o s e c A + c o t A$ , using the identity $c o s e c^{2} A = 1 + c o t^{2} A$ .

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Solution

$(v) (c o s A - s i n A + 1) / (c o s A + s i n A - 1) = c o s e c A + c o t A L . H . S . = \frac{(c o s A - s i n A + 1)}{(c o s A + s i n A - 1)} Dividing Numerator and Denominator by sin A, = \frac{\frac{(c o s A - s i n A + 1)}{s i n A}}{\frac{(c o s A + s i n A - 1)}{s i n A}} = \frac{(c o t A - 1 + c o s e c A)}{(c o t A + 1 - c o s e c A)} = \frac{(c o t A - c o s e c^{2} A + c o t^{2} A + c o s e c A)}{(c o t A + 1 - c o s e c A)} (∵ c o s e c^{2} A - c o t^{2} A = 1) = \frac{[(c o t A + c o s e c A) - (c o s e c^{2} A - c o t^{2} A)]}{(c o t A + 1 - c o s e c A)} = \frac{[(c o t A + c o s e c A) - (c o s e c A + c o t A) (c o s e c A - c o t A)]}{(1 - c o s e c A + c o t A)} = \frac{(c o t A + c o s e c A) (1 - c o s e c A + c o t A)}{(1 - c o s e c A + c o t A)} = c o t A + c o s e c A = R . H . S .$

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Similar questions

Q. Question 5 (v)
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(v)

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

, using the identity

c o s e c^{2} A = 1 + c o t^{2} A

.

Q. Question 5 (ii)
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(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

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

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

.

Q. Question 5 (ii)
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(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

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

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

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

\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

cos θ

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

[Hint: Simplify LHS and RHS separately].

\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

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

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

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

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

[Hint: Simplify LHS and RHS separately]

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

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Values of Trigonometric Ratios

Standard IX Mathematics

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