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Standard X

Mathematics

Trigonometric Ratios

Question 5x P...

Question

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

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Solution

$(i) \frac{1 + t a n^{2} A}{1 + c o t^{2} A}) = \frac{1 + t a n^{2} A}{1 + \frac{1}{t a n^{2} A}}) = \frac{1 + t a n^{2} A}{[\frac{(1 + t a n^{2} A)}{t a n^{2} A}]} = t a n^{2} A$
$(i i) (\frac{1 - t a n A}{1 - c o t A})^{2} = [\frac{1 - \frac{s i n A}{c o s A}}{1 - \frac{c o s A}{s i n A}}]^{2} = \frac{(s i n^{2} A) \times (c o s A - s i n A)^{2}}{(c o s^{2} A) \times (s i n A - c o s A)^{2}} = \frac{(s i n^{2} A)}{(c o s^{2} A)} \times (- 1)^{2} = \frac{s i n^{2} A}{c o s^{2} A} = t a n^{2} A$
$∴$ $(\frac{1 + t a n^{2} A}{1 + c o t^{2} A}) = (\frac{1 - t a n A}{1 - c o t A})^{2} = t a n^{2} A$ .

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

Q. (i) If

cos 3 A = sin (A - 34^{0})

, where A is an acute angle, find the value of A.
(ii) Prove the following identity, where the angles involved are acute angles for which the expression is define.

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

Q. Question 5 (vi)
Prove the following identities, where the angles involved are acute angles for which the expressions is defined.
sec A + tan A =

\sqrt{(\frac{1 + s i n A}{1 - s i n A})}

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

(s i n A + c o s e c A)^{2} + (c o s A + s e c A)^{2} = 7 + t a n^{2} A + c o t^{2} 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

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Question 5 (x) Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (x) (1+tan2A1+cot2A)=(1−tanA1−cotA)2=tan2A

(i) 1+tan2A1+cot2A)=1+tan2A1+1tan2A)=1+tan2A[(1+tan2A)tan2A]=tan2A (ii) (1−tanA1−cotA)2=[1−sinAcosA1−cosAsinA]2=(sin2A)×(cosA−sinA)2(cos2A)×(sinA−cosA)2=(sin2A)(cos2A)×(−1)2=sin2Acos2A=tan2A ∴ (1+tan2A1+cot2A)=(1−tanA1−cotA)2=tan2A.

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