Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

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Aptitude

TNPSC G4

10th CBSE Maths

Question 1 Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Solution:

To convert the given trigonometric ratios in terms of cot functions, use trigonometric formulas

We know that,

$cosec^{2}A – cot^{2}A$ = 1

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

Since cosec function is the inverse of sin function, it is written as

$\dfrac{1}{sin^{2}A}$ = $1 + cot^{2}A$

Now, rearrange the terms, it becomes

$sin^{2}A$ = $\dfrac{1}{(1+cot^{2}A)}$

Now, take square roots on both sides, we get

sin A = ±$\dfrac{1}{(√(1+cot^{2}A)}$

The above equation defines the sin function in terms of cot function

Now, to express sec function in terms of cot function, use this formula

$sin^{2}A$ = $\dfrac{1}{ (1+cot^{2}A)}$

Now, represent the sin function as cos function

$1 – cos^{2}A$ = $\dfrac{1}{ (1+cot^{2}A)}$

Rearrange the terms,

$cos^{2}A$ = 1 – $\dfrac{1}{(1+cot^{2}A)}$

⇒$cos^{2}A$ = $\dfrac{(1-1+cot^{2}A)}{(1+cot^{2}A)}$

Since sec function is the inverse of cos function,

⇒ $\dfrac{1}{sec^{2}A}$ = $\dfrac{cot^{2}A}{(1+cot^{2}A)}$

Take the reciprocal and square roots on both sides, we get

⇒ sec A = ±$\dfrac{\sqrt{ (1+cot^{2}A)}}{cotA}$

Now, to express tan function in terms of cot function

tan A = $\dfrac{sin A}{cos A}$ and cot A = $\dfrac{cos A}{sin A}$

Since cot function is the inverse of tan function, it is rewritten as

tan A = $\dfrac{1}{cot A}$

Additional Questions

Write all the other trigonometric ratios of ∠A in terms of sec A.	Answer
Evaluate: (i) $(sin^{2}63° + sin^{2}27°)/(cos^{2}17° + cos^{2}73°)$ (ii) sin 25° cos 65° + cos 25° sin 65°	Answer
Choose the correct option. Justify your choice. (i) $9 sec^{2}A – 9 tan^{2}A$ = (A) 1 (B) 9 (C) 8 (D) 0 (ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) (A) 0 (B) 1 (C) 2 (D) – 1 (iii) (sec A + tan A) (1 – sin A) = (A) sec A (B) sin A (C) cosec A (D) cos A (iv) $1+tan^{2}A/1+cot^{2}A$ = (A) $sec^{2} A$ (B) -1 (C) $cot^{2}A$ (D) $tan^{2}A$	Answer
Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (i) $(cosec θ – cot θ)^{2}$ = $\dfrac{(1-cos θ)}{(1+cos θ)}$ (ii) $\dfrac{cos A}{(1+sin A)}$ + $\dfrac{(1+sin A)}{cos A}$ = 2 sec A (iii) $\dfrac{tan θ}{(1-cot θ)}$ + $\dfrac{cot θ}{(1-tan θ)}$ = 1 + sec θ cosec θ [Hint : Write the expression in terms of sin θ and cos θ] (iv) $\dfrac{(1 + sec A)}{sec A}$ = $\dfrac{sin2A}{(1-cos A) }$ [Hint : Simplify LHS and RHS separately] (v) $\dfrac{( cos A–sin A+1)}{( cos A +sin A–1)}$ = cosec A + cot A, using the identity $cosec^{2}A$ = $1+cot^{2}A$. (vi)$\sqrt{\dfrac{1+sinA}{1-sinA}}$=secA+tanA (vii) $\dfrac{(sin θ – 2sin^{3}θ)}{(2cos^{3}θ-cos θ)}$ = tan θ (viii) (sin A + cosec A)$^{2}$ + (cos A + sec A)$^{2}$ = 7+tan2A+cot2A (ix) (cosec A – sin A)(sec A – cos A) = $\dfrac{1}{(tan A+cotA)}$ [Hint : Simplify LHS and RHS separately] (x) ($\dfrac{1+tan^{2}A}{1+cot^{2}A}$) = ($\dfrac{1-tan A}{1-cot A})^{2}$ = $tan^{2}A$	Answer
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.	Answer
Write all the other trigonometric ratios of ∠A in terms of sec A.	Answer
Evaluate: (i) $(sin^{2}63° + sin^{2}27°)/(cos^{2}17° + cos^{2}73°)$ (ii) sin 25° cos 65° + cos 25° sin 65°	Answer
Choose the correct option. Justify your choice. (i) $9 sec^{2}A – 9 tan^{2}A$ = (A) 1 (B) 9 (C) 8 (D) 0 (ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) (A) 0 (B) 1 (C) 2 (D) – 1 (iii) (sec A + tan A) (1 – sin A) = (A) sec A (B) sin A (C) cosec A (D) cos A (iv) $1+tan^{2}A/1+cot^{2}A$ = (A) $sec^{2} A$ (B) -1 (C) $cot^{2}A$ (D) $tan^{2}A$	Answer
Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (i) $(cosec θ – cot θ)^{2}$ = $\dfrac{(1-cos θ)}{(1+cos θ)}$ (ii) $\dfrac{cos A}{(1+sin A)}$ + $\dfrac{(1+sin A)}{cos A}$ = 2 sec A (iii) $\dfrac{tan θ}{(1-cot θ)}$ + $\dfrac{cot θ}{(1-tan θ)}$ = 1 + sec θ cosec θ [Hint : Write the expression in terms of sin θ and cos θ] (iv) $\dfrac{(1 + sec A)}{sec A}$ = $\dfrac{sin2A}{(1-cos A) }$ [Hint : Simplify LHS and RHS separately] (v) $\dfrac{( cos A–sin A+1)}{( cos A +sin A–1)}$ = cosec A + cot A, using the identity $cosec^{2}A$ = $1+cot^{2}A$. (vi)$\sqrt{\dfrac{1+sinA}{1-sinA}}$=secA+tanA (vii) $\dfrac{(sin θ – 2sin^{3}θ)}{(2cos^{3}θ-cos θ)}$ = tan θ (viii) (sin A + cosec A)$^{2}$ + (cos A + sec A)$^{2}$ = 7+tan2A+cot2A (ix) (cosec A – sin A)(sec A – cos A) = $\dfrac{1}{(tan A+cotA)}$ [Hint : Simplify LHS and RHS separately] (x) ($\dfrac{1+tan^{2}A}{1+cot^{2}A}$) = ($\dfrac{1-tan A}{1-cot A})^{2}$ = $tan^{2}A$	Answer

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