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 = $\dfrac{1}{cot A}$
Cos A function in terms of sec A:
sec A = $\dfrac{1}{cos A}$
⇒ cos A = $\dfrac{1}{sec A}$
sec A function in terms of sec A:
$cos^{2}A + sin^{2}A $= 1
Rearrange the terms
$sin^{2}A $= $1 – cos^{2}A$
$sin^{2}A$ = 1 – ($\dfrac{1}{sec^{2}A}$)
$sin^{2}A$ = $\dfrac{(sec^{2}A-1)}{sec^{2}A}$
sin A = ± $\dfrac{\sqrt{(sec^{2}A-1)}}{sec A}$
cosec A function in terms of sec A:
sin A = $\dfrac{1}{cosec A}$
⇒cosec A = $\dfrac{1}{sin A}$
cosec A = ± $\dfrac{sec A}{\sqrt(sec^{2}A-1)}$
Now, tan A function in terms of sec A:
$sec^{2}A – tan^{2}A$ = 1
Rearrange the terms
⇒ $tan^{2}A$ = $sec^{2}A – 1$
tan A = $\sqrt{(sec^{2}A – 1)}$
cot A function in terms of sec A:
tan A = $\dfrac{1}{cot A}$
⇒ cot A = $\dfrac{1}{tan A}$
cot A = ±$\dfrac{1}{\sqrt{(sec^{2}A – 1)}}$
(i) $(sin^{2}63° + sin^{2}27°)/(cos^{2}17° + cos^{2}73°)$
(ii) sin 25° cos 65° + cos 25° sin 65°
(i) $\dfrac{(sin^{2}63° + sin^{2}27°)}{(cos^{2}17° + cos^{2}73°)}$
To simplify this, convert some of the sin functions into cos functions and cos function into sin function and it becomes,
= $\dfrac{[sin^{2}(90°-27°) + sin^{2}27°] }{ [cos^{2}(90°-73°) + cos^{2}73°)]}$
= $\dfrac{(cos^{2}27° + sin^{2}27°)}{(sin^{2}27° + cos^{2}73°)}$
= $\dfrac{1}{1}$ =1 (since $sin^{2}A + cos^{2}A$ = 1)
Therefore, $\dfrac{(sin^{2}63° + sin^{2}27°)}{(cos^{2}17° + cos^{2}73°)}$ = 1
(ii) sin 25° cos 65° + cos 25° sin 65°
To simplify this, convert some of the sin functions into cos functions and cos function into sin function and it becomes,
= sin(90°-25°) cos 65° + cos (90°-65°) sin 65°
= cos 65° cos 65° + sin 65° sin 65°
= $cos^{2}65° + sin^{2}65° = 1 (since sin^{2}A + cos^{2}A = 1)$
Therefore, sin 25° cos 65° + cos 25° sin 65° = 1
(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) $tan^{2}A$
(i) (B) is correct.
Justification:
Take 9 outside, and it becomes
$9 sec^{2}A – 9 tan^{2}A$
= 9 $(sec^{2}A – tan^{2}A)$
= 9×1 = 9 (∵ $sec^{2} A – tan^{2} A$ = 1)
Therefore, 9 $sec^{2}A – 9 tan^{2}A$ = 9
(ii) (C) is correct
Justification:
(1 + tan θ + sec θ) (1 + cot θ – cosec θ)
We know that, tan θ = sin θ/cos θ
sec θ = $\dfrac{1}{ cos θ}$
cot θ = $\dfrac{cos θ}{sin θ}$
cosec θ = $\dfrac{1}{sin θ}$
Now, substitute the above values in the given problem, we get
=$(1 + \dfrac{sin θ}{cos θ} + \dfrac{1}{ cos θ}) (\dfrac{1 + cos θ}{sin θ} – \dfrac{1}{sin θ})$
Simplify the above equation,
= $\dfrac{(cos θ +sin θ+1)}{cos θ} × \dfrac{(sin θ+cos θ-1)}{sin θ}$
= $\dfrac{(cos θ+sin θ)^{2}-1^{2}}{(cos θ sin θ)}$
= $\dfrac{(cos^{2}θ + sin^{2}θ + 2cos θ sin θ -1)}{(cos θ sin θ)}$
= $\dfrac{(1+ 2cos θ sin θ -1)}{(cos θ sin θ)}$ (Since $cos^{2}θ + sin^{2}θ$ = 1)
= $\dfrac{(2cos θ sin θ)}{(cos θ sin θ)}$ = 2
Therefore, (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =2
(iii) (D) is correct.
Justification:
We know that,
Sec A= $\dfrac{1}{cos A}$
Tan A = $\dfrac{sin A }{ cos A}$
Now, substitute the above values in the given problem, we get
(secA + tanA) (1 – sinA)
=$ (\dfrac{1}{cos A} + \dfrac{sin A}{cos A}) (1 – sinA)$
= (1+$\dfrac{sin A}{cos A}$) (1 – sinA)
= $\dfrac{(1 – sin^{2}A)}{cos A}$
= $\dfrac{cos^{2}A}{cos A}$ = cos A
Therefore, (secA + tanA) (1 – sinA) = cos A
(iv) (D) is correct.
Justification:
We know that,
$tan^{2}A$ =$\dfrac{1}{cot^{2}A}$
Now, substitute this in the given problem, we get
$\dfrac{1+tan^{2}A}{1+cot^{2}A}$
= $\dfrac{(1+\dfrac{1}{cot^{2}A})}{1+cot^{2}A}$
= $(cot^{2}A+\dfrac{1}{cot^{2}A})×(\dfrac{1}{1+cot^{2}A})$
= $\dfrac{1}{cot^{2}A}$ = $tan^{2}A$
So, $\dfrac{1+tan^{2}A}{1+cot^{2}A}$ = $tan^{2}A$
(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) ($\dfrac{1+tan^{2}A}{1+cot^{2}A}$) = ($\dfrac{1-tan A}{1-cot A})^{2}$ = $tan^{2}A$
(i) $(cosec θ – cot θ)^{2}$ = $\dfrac{(1-cos θ)}{(1+cos θ)}$
To prove this, first take the Left-Hand side (L.H.S) of the given equation, to prove the Right Hand Side (R.H.S)
L.H.S. = $(cosec θ – cot θ)^{2}$
The above equation is in the form of $(a-b)^{2}$, and expand it
Since $(a-b)^{2}$ = a$^{2}$ + b$^{2}$ – 2ab
Here a = cosec θ and b = cot θ
= $(cosec^{2}θ + cot^{2}θ – 2cosec θ cot θ)$
Now, apply the corresponding inverse functions and equivalent ratios to simplify
= $(\dfrac{1}{sin^{2}θ} + \dfrac{cos^{2}θ}{sin^{2}θ} – \dfrac{2cos θ}{sin^{2}θ}$
= $\dfrac{(1 + cos^{2}θ – 2cos θ)}{(1 – cos^{2}θ)}$
= $\dfrac{(1-cos θ)^{2}}{(1 – cosθ)(1+cos θ)}$
= $\dfrac{(1-cos θ)}{(1+cos θ)}$ = R.H.S.
Therefore,$ (cosec θ – cot θ)^{2}$ = $\dfrac{(1-cos θ)}{(1+cos θ)}$
Hence proved.
(ii)$ (\dfrac{cos A}{(1+sin A)} + \dfrac{(1+sin A)}{cos A} $= 2 sec A
Now, take the L.H.S of the given equation.
L.H.S. = $(\dfrac{cos A}{(1+sin A)}) + (\dfrac{(1+sin A)}{cos A})$
=$ [cos^{2}A + \dfrac{(1+sin A)^{2}]}{(1+sin A)cos A}$
= $\dfrac{(cos^{2}A + sin^{2}A + 1 + 2sin A)}{(1+sin A) cos A}$
Since $cos^{2}A + sin^{2}A$ = 1, we can write it as
= $\dfrac{(1 + 1 + 2sin A)}{(1+sin A) cos A}$
= $\dfrac{(2+ 2sin A)}{(1+sin A)cos A}$
= $\dfrac{2(1+sin A)}{(1+sin A)cos A}$
= $\dfrac{2}{cos A}$ = 2 sec A = R.H.S.
L.H.S. = R.H.S.
($\dfrac{cos A}{(1+sin A)}) + (\dfrac{(1+sin A)}{cos A)}$ = 2 sec A
Hence proved.
(iii) $\dfrac{tan θ}{(1-cot θ)}$ + $\dfrac{cot θ}{(1-tan θ) }$= 1 + sec θ cosec θ
L.H.S. =$\dfrac{ tan θ}{(1-cot θ)} + \dfrac{cot θ}{(1-tan θ)}$
We know that tan θ =$\dfrac{sin θ}{cos θ}$
cot θ = $\dfrac{cos θ}{sin θ}$
Now, substitute it in the given equation, to convert it in a simplified form
=$ [\dfrac{(\dfrac{sin θ}{cos θ})}{1-(\dfrac{cos θ}{sin θ}}] + [\dfrac{(\dfrac{cos θ}{sin θ})}{1-(\dfrac{sin θ}{cos θ})}]$
= [$\dfrac{(\dfrac{sin θ}{cos θ})}{\dfrac{(sin θ-cos θ)}{sin θ}}] + [\dfrac{(\dfrac{cos θ}{sin θ}}{\dfrac{(cos θ-sin θ)}{cos θ}}]$
= $\dfrac{sin^{2}θ}{[cos θ(sin θ-cos θ)]} +\dfrac {cos^{2}θ}{[sin θ(cos θ-sin θ)]}$
= $\dfrac{sin^{2}θ}{[cos θ(sin θ-cos θ)]} – \dfrac{cos^{2}θ}{[sin θ(sin θ-cos θ)]}$
= $\dfrac{1}{(sin θ-cos θ) [(sin^{2}θ/cos θ)} – (\dfrac{cos^{2}θ}{sin θ)}]$
= $\dfrac{1}{(sin θ-cos θ)} × [\dfrac{(sin^{3}θ – cos^{3}θ)}{sin θ cos θ}]$
= [$\dfrac{(sin θ-cos θ)(sin^{2}θ+cos^{2}θ+sin θ cos θ)]}{[(sin θ-cos θ)sin θ cos θ]}$
= $\dfrac{(1 + sin θ cos θ)}{sin θ cos θ}$
= $\dfrac{1}{sin θ cos θ}$ + 1
= 1 + sec θ cosec θ = R.H.S.
Therefore, L.H.S. = R.H.S.
Hence proved
(iv) $\dfrac{(1 + sec A)}{sec A}$ = $\dfrac{sin^{2}A}{(1-cos A)}$
First find the simplified form of L.H.S
L.H.S. = $\dfrac{(1 + sec A)}{sec A}$
Since secant function is the inverse function of cos function and it is written as
=$\dfrac{ (1 + \dfrac{1}{cos A})}{\dfrac{1}{cos A}}$
= $\dfrac{\dfrac{(cos A + 1)}{cos A}}{\dfrac{1}{cos A}}$
Therefore, $\dfrac{(1 + sec A)}{sec A}$ = cos A + 1
R.H.S. = $\dfrac{sin^{2}A}{(1-cos A)}$
We know that $sin^{2}A$ = $(1 – cos^{2}A)$, we get
= $\dfrac{(1 – cos^{2}A)}{(1-cos A)}$
= $\dfrac{(1-cos A)(1+cos A)}{(1-cos A)}$
Therefore, $\dfrac{sin^{2}A}{(1-cos A)}$= cos A + 1
L.H.S. = R.H.S.
Hence proved
(v) $\dfrac{(cos A–sin A+1)}{(cos A+sin A–1)}$ = cosec A + cot A, using the identity $cosec^{2}A $= $1+cot^{2}A$.
With the help of identity function, $cosec^{2}A$ = $1+cot^{2}A$, let us prove the above equation.
L.H.S. = $\dfrac{(cos A–sin A+1)}{(cos A+sin A–1)}$
Divide the numerator and denominator by sin A, we get
= $\dfrac{\dfrac{(cos A–sin A+1)}{sin A}}{\dfrac{(cos A+sin A–1)}{sin A}}$
We know that $\dfrac{cos A}{sin A}$ = cot A and $\dfrac{1}{sin A}$ = cosec A
= $\dfrac{(cot A – 1 + cosec A)}{(cot A+ 1 – cosec A)}$
= $\dfrac{(cot A – cosec^{2}A + cot^{2}A + cosec A)}{(cot A+ 1 – cosec A)}$ (using $cosec^{2}A – cot^{2}A$ = 1
=$\dfrac{ [(cot A + cosec A) – (cosec^{2}A – cot^{2}A)]}{(cot A+ 1 – cosec A)}$
= $\dfrac{[(cot A + cosec A) – (cosec A + cot A)(cosec A – cot A)]}{(1 – cosec A + cot A)}$
= $\dfrac{(cot A + cosec A)(1 – cosec A + cot A)}{(1 – cosec A + cot A)}$
= cot A + cosec A = R.H.S.
Therefore, $\dfrac{(cos A–sin A+1)}{(cos A+sin A–1)}$ = cosec A + cot A
Hence Proved
(vi)$\sqrt{\dfrac{1+sinA}{1-sinA}}$=secA+tanA
L.H.S=$\sqrt{\dfrac{1+sinA}{1-sinA}}$First divide the numerator and denominator of L.H.S. by cos A,
=$\sqrt{\dfrac{\dfrac{1}{cos A}+\dfrac{sin A}{cos A}}{\dfrac{1}{cos A}-\dfrac{sin A}{cos A}}}$
We know that 1/cos A = sec A and sin A/ cos A = tan A and it becomes,
= $\dfrac{\sqrt(sec A+ tan A)}{(sec A-tan A)}$
Now using rationalization, we get
=$\sqrt{\dfrac{sec A+tan A}{sec A-tan A}}$x$\sqrt{\dfrac{sec A+tan A}{sec A+tan A}}$
=$\sqrt{\dfrac{(sec A+tan A)^{2}}{sec^{2}A-tan^{2}A}}$
=$\dfrac{ (sec A + tan A)}{1}$
= sec A + tan A = R.H.S
Hence proved
(vii) $\dfrac{ (sin θ – 2sin^{3}θ)}{(2cos^{3}θ-cos θ)}$ = tan θ
L.H.S. = $\dfrac{ (sin θ – 2sin^{3}θ)}{(2cos^{3}θ – cos θ)}$
Take sin θ as in numerator and cos θ in denominator as outside, it becomes
= $\dfrac{ [sin θ(1 – 2sin^{2}θ)]}{[cos θ(2cos^{2}θ- 1)]}$
We know that $sin^{2}θ$ = $1-cos^{2}θ$
= $\dfrac{sin θ[1 – 2(1-cos^{2}θ)]}{[cos θ(2cos^{2}θ -1)]}$
= $\dfrac{[sin θ(2cos^{2}θ -1)]}{[cos θ(2cos^{2}θ -1)]}$
= tan θ = R.H.S.
Hence proved
(viii) $(sin A + cosec A)^{2} + (cos A + sec A)^{2} $= $7+tan^{2}A+cot^{2}A$
L.H.S. = $(sin A + cosec A)^{2} + (cos A + sec A)^{2}$
It is of the form $(a+b)^{2}$, expand it
$(a+b)^{2}$=$a^{2} + b^{2} +2ab$
= $(sin^{2}A + cosec^{2}A + 2 sin A cosec A) + (cos^{2}A + sec^{2}A + 2 cos A sec A)$
= $(sin^{2}A + cos^{2}A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan^{2}A + 1 + cot^{2}A$
= $1 + 2 + 2 + 2 + tan^{2}A + cot^{2}A$
= $7+tan^{2}A+cot^{2}A $= R.H.S.
Therefore, $(sin A + cosec A)^{2} + (cos A + sec A)^{2}$ = $7+tan^{2}A+cot^{2}A$
Hence proved.
(ix) (cosec A – sin A)(sec A – cos A) = $\dfrac{1}{(tan A+cotA)}$
First, find the simplified form of L.H.S
L.H.S. = (cosec A – sin A)(sec A – cos A)
Now, substitute the inverse and equivalent trigonometric ratio forms
= $(\dfrac{1}{sin A – sin A})(\dfrac{1}{cos A – cos A})$
=$ [\dfrac{(1-sin^{2}A)}{sin A}][\dfrac{(1-cos^{2}A)} {cos A}]$
= $(\dfrac{cos^{2}A}{sin A})×(\dfrac{sin^{2}A}{cos A})$
= cos A sin A
Now, simplify the R.H.S
R.H.S. =$\dfrac{ 1}{(tan A+cotA)}$
= $\dfrac{1}{(\dfrac{sin A}{cos A} +\dfrac{cos A}{sin A})}$
= $\dfrac{1}{[(sin^{2}A+\dfrac{cos^{2}A)}{sin A cos A}]}$
= cos A sin A
L.H.S. = R.H.S.
(cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
Hence proved
(x) $(\dfrac{1+tan^{2}A}{1+cot^{2}A})$ = $(\dfrac{1-tan A}{1-cot A})^{2} = tan^{2}A$
L.H.S. = $(\dfrac{1+tan^{2}A}{1+cot^{2}A})$
Since cot function is the inverse of tan function,
= $(\dfrac{1+tan^{2}A}{1+\dfrac{1}{tan^{2}A}})$
= $\dfrac{1+tan^{2}A}{[\dfrac{(1+tan^{2}A)}{tan^{2}A]}}$
Now cancel the $1+tan^{2}A$ terms, we get
= $tan^{2}A$
$(\dfrac{1+tan^{2}A}{1+cot^{2}A})$ = $tan^{2}A$
Similarly,
$(\dfrac{1-tan A}{1-cot A})^{2}$ = $tan^{2}A$
Hence proved