Revision: Trigonometry JEE Main Trigonometry

When an equation, involving trigonometrical ratios of an angle A, is true for all values of A, the equation is called a trigonometric identity. 

The straight line joining the eye of the observer to the point on the object being viewed.

The angle between the line of sight and the horizontal through the observer’s eye, when the object is below the level of the observer’s eye.

The angle between the line of sight and the horizontal through the observer’s eye, when the object is above the level of the observer’s eye.

\[sineA=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\]

\[cosineA=\frac{\mathrm{Base}}{\text{Hypotenuse}}\]

\[tangentA=\frac{\text{Perpendicular}}{\mathrm{Base}}\]

\[cotangent A = \frac{\text{Base}}{\text{Perpendicular}}\]

\[secantA=\frac{\text{Hypotenuse}}{\mathrm{Base}}\]

\[cosecantA=\frac{\text{Hypotenuse}}{\text{Perpendicular}}\]

L.H.S. = `sqrt(((1 - sin θ)(1 - sin θ))/((1 + sin θ)(1 - sin θ)))`

= `sqrt((1 + sin^2θ - 2sinθ)/(1 - sin^2θ)`

= `sqrt((1 + sin^2θ - 2sinθ)/(cos^2θ)`

= `sqrt( 1/cos^2θ + sin^2θ/cos^2θ - (2sin θ)/cos θ xx 1/cosθ`

= `sqrt( sec^2θ + tan^2 θ - 2 tan θ . sec θ)`

= `sqrt((sec θ - tan θ)^2)`

= sec θ – tan θ

= R.H.S.

Hence proved.

L.H.S. = `sqrt((1 - sin θ)/(1 + sin θ))`

= `sqrt(((1 - sin θ)(1 - sin θ))/((1 + sin θ)(1 - sin θ))`

= `sqrt(((1 - sin θ)^2)/(1 - sin^2θ)`

= `sqrt(((1 - sin θ)^2)/(cos^2θ)`

= `(1 - sin θ)/(cos θ)`

= `1/(cos θ) - (sin θ)/(cos θ)`

= sec θ – tan θ

= R.H.S.

Hence Proved.

L.H.S. = `secA/(secA + 1) + secA/(secA - 1)`

= `(sec^2A - secA + sec^2A + secA)/(sec^2A - 1`

= `(2sec^2A)/tan^2A`   ...(∵ sec² A – 1 = tan² A)

= `(2/cos^2A)/(sin^2A/cos^2A)`

= `2/sin^2A`

= 2 cosec² A = R.H.S.

We know that `sin^2 theta + cos^2 theta = 1`

So,

LHS = `"cosec" theta sqrt(1 - cos^2 theta)`

= `"cosec" theta sqrt (sin^2 theta)`

= cosec θ . sin θ

`1/sin theta xx sin theta`

= 1

= RHS hence proved.

LHS = `1 + (tan^2 θ)/((1 + sec θ))`

=` 1 + ((sec^2 θ - 1))/((sec theta + 1))`

=`1 + ((sec theta + 1)(sec theta - 1))/((sec theta + 1))`

=`1 + (sec theta - 1)`

= sec θ

LHS = RHS

LHS =`1/((1+ sin θ)) + 1/((1 - sin θ))`

= `((1 - sin θ) + (1 + sin θ))/((1 + sin θ)(1 - sin θ))`

= `2/(1 - sin^2 θ)`

= `2/(cos^2 θ)`

= 2 sec² θ

= RHS

Hence Proved.

LHS = `sqrt((1 + sin θ)/(1 - sin θ))`

=`sqrt(((1 + sin θ))/(1 - sin θ) xx ((1 + sin θ))/(1 + sin θ))`

=` sqrt(((1 + sin θ)^2)/(1 - sin^2 θ))`

=`sqrt(((1 + sin θ)^2)/(cos^2 θ))`

=`(1 + sin θ)/cos θ`

=`1/cos θ + (sin θ)/(cos θ)`

= sec θ + tan θ

= RHS

LHS = `(sin θ - 2sin^3 θ)/(2 cos^3 θ - cos θ)`

= `(sin θ(1 - 2sin^2 θ))/(cos θ(2 cos^2 θ - 1))`

= `(tan θ(1 - 2(1 - cos^2 θ)))/(2 cos^2θ - 1 )`

= `(tan θ(1 - 2 + 2 cos^2 θ))/(2 cos^2θ - 1 )`

= `(tan θ(2 cos^2 θ - 1))/(2 cos^2θ - 1 )`

= tan θ

= RHS

Hence proved.

L.H.S. = `tanA/(1 - cotA) + cotA/(1 - tanA)`

= `tanA/(1 - 1/tanA) + (1/tanA)/(1 - tanA)`

= `tan^2A/(tanA - 1) + 1/(tanA(1 - tanA))`

= `(tan^3A - 1)/(tanA(1 - tanA))`

= `((tanA - 1)(tan^2A + 1 + tanA))/(tanA(tanA - 1)`

= `(sec^2A + tanA)/tanA`

= `(1/cos^2A)/(sinA/cosA) + 1`

= `1/(sinAcosA) + 1`

= sec A cosec A + 1 = R.H.S.

LHS = `sqrt((1 - cos θ)/(1 + cos θ) xx (1 - cos θ)/(1 - cos θ))`

= `sqrt((1 - cos θ)^2/(1 - cos^2θ))`

= `(1 - cos θ)/(sqrt(1 - cos^2θ))` 

= `(1 - cos θ)/(sqrt(sin^2θ))`

= `(1 - cos θ)/(sin θ)`

= `(1)/(sin θ) - (cos θ)/(sin θ)`

= cosec θ − cot θ
= RHS
Hence proved.

Given that, tan A = n tan B and sin A = m sin B.

`=> n = tanA/tanB` and `m = sinA/sinB` 

∴ `(m^2 - 1)/(n^2 - 1) = ((sinA/sinB)^2 - 1)/((tanA/tanB)^2 - 1)`

= `(sin^2A/sin^2B - 1/1)/(tan^2A/(tan^2B) - 1)`

= `((sin^2A - sin^2B).tan^2B)/(sin^2B.(tan^2A - tan^2B))`

= `((sin^2A - sin^2B)/tan^2B)/((tan^2A - tan^2B)/sin^2B)`

= `((sin^2A - sin^2B)sin^2B)/((sin^2A/cos^2A-sin^2B/cos^2B)cos^2Bsin^2B)`

= `(sin^2A - sin^2B)/(((sin^2A.cos^2B - sin^2B.cos^2A)/(cos^2A.cos^2B)) cos^2B)`

= `((sin^2A - sin^2B)cos^2A)/(sin^2A.cos^2B - sin^2B.cos^2A)`

= `((sin^2A - sin^2B)cos^2A)/(sin^2A(1 - sin^2B) - sin^2B (1 - sin^2A))`

= `((sin^2A - sin^2B)cos^2A)/(sin^2A - sin^2A.sin^2B - sin^2B + sin^2B.sin^2A)`

= `((sin^2A - sin^2B)cos^2A)/(sin^2A -sin^2B)`

= cos² A

We have to prove `(1 + sin θ)/cos θ + cos θ/(1 + sin θ) = 2 sec θ`

We know that, `sin^2 θ + cos^2 θ = 1`

Multiplying the denominator and numerator of the second term by (1 − sin θ), we have

= `(1 + sin θ)/cos θ + cos θ/(1 + sin θ)`

`(1 + sin θ)/cos θ =  (cos θ(1 - sin θ))/((1 + sin θ)(1 - sin θ))`

`(1 + sin θ)/cos θ =  (cos θ (1 - sin θ))/(1-sin θ)`

= `(1 + sin θ)/cos θ + (cos θ(1 - sin θ))/cos^2 θ`

= `(1 + sin θ)/cos θ + (1 - sin θ)/cos θ`

= `(1 + sin θ +  1 - sin θ)/cos θ`

`= 2/cos θ`

= 2 sec θ

LHS = `(1 + sin θ)/cos θ + cos θ/(1 + sin θ)`

= `(( 1 + sin θ)^2 + cos^2 θ)/(cos θ(1 + sin θ))`

= `(1 + sin^2 θ + 2 sin θ + cos^2 θ)/(cos θ(1 + sin θ ))`

= `(1 + (sin^2θ + cos^2 θ) + 2 sin θ)/(cos θ(1 + sin θ))`

= `(1 + 1 + 2sin θ)/(cos θ(1 + sin θ))`

= `(2(1 + sin θ))/(cos θ(1 + sin θ))`

= 2 sec θ

Hence proved.

We know that `sec^2 theta - tan^2 theta = 1`

So,

`tan theta + 1/tan theta = (tan^2 theta + 1)/tan theta`

`= sec^2 theta/tan theta`

`= sec theta sec theta/tan theta`

`= sec theta = (1/cos theta)/(sin theta/cos theta)`

`= sec theta cosec theta`

LHS = `(sec θ - tan θ)/(sec θ + tan θ )`

= `(sec θ - tan θ)/(sec θ + tan θ ) xx (sec θ - tan θ)/(sec θ - tan θ )`

= `(sec θ - tan θ)^2/(sec^2θ - tan^2θ )`

= `(sec^2θ + tan^2θ - 2sec θ.tan θ )/1`

= 1 + 2 tan²θ − 2 sec θ. tan θ

= R.H.S.
Hence proved.

In the given question, we need to prove `1/(sec A + tan A) - 1/cos A = 1/cos A - 1/(sec A - tan A)`

Here, we will first solve the L.H.S.

Now using `sec theta = 1/cos theta` and `tan theta = sin theta/cos theta`, we get

`1/(sec A +  tan A) - 1/cos A  = 1/(1/cos A + sin A/cos A) - (1/cos A)`

`= 1/(((1 + sin A)/cos A)) - (1/cos A)`

`= (cos A/(1 + sin A)) - (1/cos A)`

`= (cos^2 A - (1 + sin A))/((1 + sin A)(cos A))`

On further solving, we get

`(cos^2 A -(1 + sin A))/((1 + sin A)(cos A)) = (cos^2 A - 1 - sin A)/((1 +  sin A)(cos A))`

`= (-sin^2 A - sin A)/((1 + sin A)(cos A))`    (Using `sin^2 theta = 1 - cos^2 theta)`

`= (-sin A(sin A + 1))/((1 + sin A)(cos A))`

`= (-sin A)/cos A`

= − tan A

Similarly, we solve the R.H.S.

`((1 - sin A) - cos^2 A)/((cos A)(1 - sin^2 A)) = (1 - sin A - cos^2 A)/((cos A)(1 - sin A))`

`= (sin^2 A - sin A)/((cos A)(1 - sin A))`   (Using `sin^2 theta = 1- cos^2 theta`) 

`= (-sin A(1 - sin A))/((cos A)(1 - sin A))`

`= (-sin A)/cos A`

= − tan A

So, L.H.S = R.H.S

Hence proved.

We know that

sec² θ − tan² θ = 1

cosec² θ − cot² θ = 1

So,

(sec² θ − 1)(cosec² θ − 1) = tan² θ × cot² θ

= (tan θ × cot θ)

= `(tan θ xx 1/tan θ)^2`

= (1)²

= 1

We need to prove `sec^6 theta = tan^6 theta + 3 tan^2 theta sec^2 theta + 1`

Solving the L.H.S, we get

`sec^6 theta = (sec^2 theta)^3`

`= (1 + tan^2 theta)^3`

Further using the identity `(a + b)^3 = a^3 + b^3 + 3a^2b + 3ab^2`, we get

`(1 + tan^2 theta)^3 = 1 + tan^6 theta + 3(1)^2 (tan^2 theta) + 3(1)(tan^2 theta)^2`

`= 1 + tan^6 theta + 3 tan^2 theta + 3 tan^4 theta`

`= 1 + tan^6 theta + 3 tan^2 theta + 3 tan^4 theta`

`= 1 + tan^6 theta + 3 tan^2 theta (1 + tan^2 theta)`

`= 1 + tan^6 theta + 3 tan^2 theta sec^2 theta`   (using `1 + tan^2 theta = sec^2 theta`)

Hence proved.

We need to prove `tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = 1 + tan θ + cot θ`

Now using cot θ = `1/tan θ` in the LHS, we get

`tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = tan θ/(1 - 1/tan θ) + (1/tan θ)/(1 - tan θ)`

`= tan θ/(((tan θ - 1)/tan θ)) + 1/(tan θ(1 - tan θ))`

`= (tan θ)/(tan θ  - 1)(tan θ) + 1/(tan θ(1 - tan θ)`

`= tan^2 θ/(tan θ - 1) - 1/(tan θ(tan θ - 1))`

`= (tan^3 θ - 1)/(tan θ(tan θ - 1))`

Further using the identity `a^3 - b^3 = (a - b)(a^2 + ab + b^2)`, we get

`(tan^3 θ - 1)/(tan(tan θ - 1)) = ((tan θ - 1)(tan^2 θ + tan θ + 1))/(tan θ (tan θ - 1))`

`= (tan^2 θ + tan θ + 1)/(tan θ)`

`= tan^2 θ/tan θ+ tan θ/tan θ + 1/tan θ`

= tan θ + 1 + cot θ

Hence `tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = 1 + tan θ + cot θ`

`(1 + cos θ + sin θ)/(1 + cos θ - sin θ) = (1 + sin θ)/cos θ`

Consider the LHS = `(1 + cos θ + sin θ)/(1 + cos θ - sin θ)`

`= ((1 + cos θ + sin θ)/(1 + cos θ - sin θ))((1 + cos θ + sin θ)/(1 + cos θ + sin θ))`

`= (1 + cos θ + sin θ)^2/((1 + cos θ)^2 sin^2 θ)`

`= (2 + 2(cos θ + sin θ + sin θ cos θ))/(2 cos^2 θ+ 2 cos θ)`

`= (2(1 + cos θ)(1 + sin θ))/(2 cos θ (1 + cos θ))`

`= (1 + sin θ)/cos θ`

= RHS

Hence proved

LHS = `(sin θ/cos θ + sin θ)/(sin θ/cos θ - sin θ)`

= `(sin θ (1/cos θ + 1))/(sin θ (1/cos θ - 1))`

= `(sec θ + 1)/(sec θ - 1)`

= RHS

Hence proved.

Given: 1 + sin² θ = 3 sin θ cos θ

Dividing L.H.S and R.H.S equations with sin²θ,

We get, 

`(1 + sin^2 theta)/(sin^2 theta) = (3 sin theta cos theta)/(sin^2 theta)`

`\implies 1/(sin^2 theta) + 1 = (3 cos theta)/sintheta`

cosec² θ + 1 = 3 cot θ

Since, cosec² θ – cot² θ = 1 

`\implies` cosec² θ = cot² θ + 1

`\implies` cot² θ + 1 + 1 = 3 cot θ

`\implies` cot² θ + 2 = 3 cot θ

`\implies` cot² θ – 3 cot θ + 2 = 0

Splitting the middle term and then solving the equation,

`\implies` cot² θ – cot θ – 2 cot θ + 2 = 0

`\implies` cot θ(cot θ – 1) – 2(cot θ + 1) = 0

`\implies` (cot θ – 1)(cot θ – 2) = 0

`\implies` cot θ = 1, 2

Since,

tan θ = `1/cot θ`

tan θ = `1, 1/2`

Hence proved.

Given, 1 + sin² θ = 3 sin θ cos θ

On dividing by sin² θ on both sides, we get

`1/(sin^2θ) + 1 = 3 cot θ`   ...`[∵ cot θ = cos θ/sin θ]`

⇒ cosec² θ + 1 = 3 cot θ

⇒ 1 + cot² θ + 1 = 3 cot θ

⇒ cot² θ – 3 cot θ + 2 = 0

⇒ cot² θ – 2 cot θ – cot θ + 2 = 0

⇒ cot θ (cot θ – 2) – 1(cot θ – 2) = 0

⇒ (cot θ – 2) (cot θ – 1) = 0

⇒ cot θ = 1 or 2

tan θ = 1 or `1/2`

Hence proved.

sin θ + cos θ = `sqrt(3)`

Squaring on both sides:

(sin θ + cos θ)² = `(sqrt(3))^2`

sin² θ + cos² θ + 2 sin θ cos θ = 3

1 + 2 sin θ cos θ = 3

2 sin θ cos θ = 3 – 1

2 sin θ cos θ = 2

∴ sin θ cos θ = 1

L.H.S = tan θ + cot θ

= `sin theta/cos theta + cos theta/sin theta`

= `(sin^2 theta + cos^2 theta)/(sin theta cos theta)`

= `1/(sin theta cos theta)`

= `1/1`   ...(sin θ cos θ = 1)

= 1 = R.H.S.

⇒ tan θ + cot θ = 1

L.H.S = R.H.S

LHS = (sin θ + cos θ)(cosec θ – sec θ)

= `(sin θ + cos θ)(1/sin θ - 1/cos θ)`

= `(sin θ + cos θ)((cos θ - sin θ)/(sin θ * cos θ))`

= `(cos^2θ - sin^2θ)/(sinθ * cosθ)`

= `(1 - 2sin^2θ)/(sinθ*cosθ)`

= `1/(sinθ * cosθ) - (2 sin^2θ)/(sinθ * cosθ)`

= `cosec θ · sec θ - (2 sin^2 θ)/(sin θ * cos θ)`

= cosec θ · sec θ – 2 tan θ

= RHS

Hence proved.

Given that: m sinθ = n sin(θ + 2α)

⇒ `(sin(theta + 2alpha))/sintheta = m/n`

Using componendo and dividendo theorem, we get,

⇒ `(sin(theta + 2alpha) + sintheta)/(sin(theta + 2alpha) - sintheta) = (m + n)/(m - n)`

⇒ `(2sin((theta + 2alpha + theta)/2).cos((theta + 2alpha - theta)/2))/(2cos((theta + 2alpha + theta)/2).sin((theta + 2alpha - theta)/2)) = (m + n)/(m - n)`    .......`[(because sinA + sinB = 2sin  (A + B)/2 . cos  (A - B)/2),(sinA - sinB = 2cos  (A + B)/2 . sin  (A - B)/2)]`

⇒ `(sin(theta + alpha).cosalpha)/(cos(theta + alpha).sinalpha) = (m + n)/(m - n)`

⇒ `tan(theta + alpha)cotalpha = (m + n)/(m - n)`

Hence proved.

Given that: cosα + cosβ = 0

And sinα + sinβ = 0

∴ (cosα + cosβ)² - (sinα + sinβ)² = 0

⇒ (cos²α + cos²β + 2cosα cosβ) - (sin²α + sin²β + 2sinα sinβ) = 0

⇒ cos²α + cos²β + 2 cosα cosβ - sin²α - sin²β - 2 sinα sinβ = 0

⇒ (cos²α - sin²α) + (cos²β - sin²β) + 2(cosα cosβ - sinα sinβ) = 0

⇒ cos2α + cos2β + 2cos(α + β) = 0

Hence, cos2α + cos2β = -2cos(α + β).

Hence proved

Given that: cos(α + β) = `4/5`

∴ tan(α + β) = `3/4`

And sin(α – β) = `5/13`

∴ tan(α – β) = `5/12`

Now tan 2α = tan[α + β + α – β]

= tan[(α + β) + (α – β)]

= `(tan(alpha + beta) + tan(alpha - beta))/(1 - tan(alpha + beta).tan(alpha - beta))`

= `(3/4 + 5/12)/(1 - 3/4 xx 5/12)`

= `((9 + 5)/12)/((48 - 15)/48)`

= `14/12 xx 48/33`

= `56/33`

Hence, tan 2α = `56/33`.

LHS = `(cos A)/a + (cos B)/b + (cos C)/c`

`= ((("b"^2 + "c"^2 - "a"^2)/"2bc"))/"a" + ((("c"^2 + "a"^2 - "b"^2)/"2ca"))/"b" + ((("a"^2 + "b"^2 - "c"^2)/"2ab"))/"c"`

`= ("b"^2 + "c"^2 - "a"^2)/"2abc" + ("c"^2 + "a"^2 - "b"^2)/"2abc" + ("a"^2 + "b"^2 - "c"^2)/"2abc"`

`= ("b"^2 + "c"^2 - "a"^2 + "c"^2 + "a"^2 - "b"^2 + "a"^2 + "b"^2 - "c"^2)/"2abc"`

`= ("a"^2 + "b"^2 + "c"^2)/"2abc"`

= RHS

LHS = `(cos A)/a + (cos B)/b + (cos C)/c`

= `(b cos A + a cos B)/(ab) + (cos C)/c`

= `c/(ab) + (cos C)/c`    ...(By projection rule)

= `c/(ab) + (a^2 + b^2 - c^2)/(2 abc)`    ...(By cosine rule)

= `(2c^2 + a^2 + b^2 - c^2)/(2 abc)`

= `(a^2 + b^2 + c^2)/(2 abc)` = R.H.S.

Put x = cos θ

∴ θ = cos^–1 x

L.H.S. = `tan^-1 ((sqrt(1 + x) - sqrt(1 - x))/(sqrt(1 + x) + sqrt(1 - x)))`

= `tan^-1 ((sqrt(1 + cos θ) - sqrt(1 - cos θ))/(sqrt(1 + cos θ) + sqrt(1 - cos θ)))`

= `tan^-1 [(sqrt(2 cos^2(θ/2)) - sqrt(2 sin^2 (θ/2)))/(sqrt(2 cos^2 (θ/2)) + sqrt(2 sin^2 (θ/2)))]`

= `tan^-1 [(sqrt(2) cos (θ/2) - sqrt(2) sin (θ/2))/(sqrt(2) cos (θ/2) + sqrt(2) sin (θ/2))]`

= `tan^-1 [((sqrt(2) cos (θ/2))/(sqrt(2) cos (θ/2)) - (sqrt(2) sin (θ/2))/(sqrt(2) cos (θ/2)))/((sqrt(2) cos (θ/2))/(sqrt(2) cos (θ/2)) + (sqrt(2) sin (θ/2))/(sqrt(2) cos (θ/2)))]`

= `tan^-1 [(1 - tan(θ/2))/(1 + tan (θ/2))]`

= `tan^-1 [(tan  pi/4 - tan (θ/2))/(1 + tan  pi/4. tan (θ/2))]  ....[∵ tan  pi/4 =1]`

= `tan^-1 [tan (pi/4 - θ/2)]`

= `pi/4 - θ/2`

= `pi/4 - 1/2 cos^-1`x  .....[∵ θ = cos^–1 x]

∴ L.H.S. = R.H.S.

Let `cos^(-1)  4/5` = x

Then, cos x = `4/5`

⇒ sin x = `sqrt (1 - (4/5)^2)`

⇒ sin x = `sqrt(1 - 16/25)`

⇒ sin x = `sqrt(9/25)`

⇒ sin x = `3/5`

∴ tan x = `3/4` ⇒ x = `tan^(-1)  3/4`

∴ `cos^(-1)  4/5 =  tan^(-1)  3/4`   ...(1)

Now let `cos^(-1)  12/13` = y

Then cos y = `12/13`

⇒ sin y = `5/13`

∴ tan y = `5/12` ⇒ y = `tan^(-1)  5/12`

∴ `cos^(-1)  12/13 = tan^(-1)  5/12`  ....(2)

Let `cos^(-1)  33/65` = z

Then cos z = `33/65`

⇒ sin z = `56/65`

∴ tan z = `56/33` ⇒ z = `tan^(-1)  56/33`

∴ `cos^(-1)  33/65 = tan^(-1)  56/33`  ....(3)

Now, we will prove that:

L.H.S = `cos^(-1)  4/5 + cos^(-1)  12/13`

= `tan^(-1)  3/4 + tan^(-1)  5/12`  ....[Using (1) and (2)]

= `tan^(-1)  (3/4 + 5/12)/(1 - 3/4 * 5/12)    ....[tan^(-1) x + tan^(-1) y = tan^(-1)  (x + y)/(1 - xy)]`

= `tan^(-1)  (36+20)/(48-15)`

= `tan^(-1)  56/33`

= `cos^(-1)  33/65`   .....[by (3)]

= R.H.S.

Let x = cos θ

Then, cos⁻¹x =  θ

We have,

R.H.S = cos⁻¹(4x³ − 3x)

⇒ cos⁻¹(4 cos³θ − 3 cos θ)

⇒ cos⁻¹(cos 3θ) = cos⁻¹(4x³ − 3x)

⇒ 3θ =  cos⁻¹(4x³ − 3x)

⇒ 3 cos⁻¹x = cos⁻¹(4x³ − 3x)

R.H.S = L.H.S

`sin^-1  8/17 + sin^-1  3/5`

= `tan^-1  8/sqrt(17^2 - 8^2) + tan^-1  3/sqrt(5^2 - 3^2)  ...[sin^-1  p/h = tan^-1  p/sqrt(h^2 - p^2)]`

= `tan^-1  8/sqrt(289 - 64) + tan^-1  3/sqrt(25 - 9)`

= `tan^-1  8/sqrt225 + tan^-1  3/sqrt16`

= `tan^-1  8/15 + tan^-1  3/4`

= `tan^-1  ((8/15 + 3/4)/(1 - 8/15 xx 3/4))  ...[tan^-1x + tan^-1y = tan^-1((x + y)/(1 - x xx y))]`

= `tan^-1[((32 + 45)/60)/(1 - 24/60)]`

= `tan^-1  77/36`

Let x = `cos^(-1)  12/13` and y = `sin^(-1)  3/5`

or cos x = `12/13` and sin y = `3/5`

sin x = `sqrt (1 - cos^2 x)` and cos y = `sqrt(1 - sin^2 y)`

Now, sin x = `sqrt(1 - 144/169)` and cos y = `sqrt( 1 - 9/25)`

⇒ sin x = `5/13` and cos y = `4/5`

We know that,

sin (x + y) = sin x cos y + cos x sin y

= `5/13 xx 4/5 + 12/13 xx 3/5 `

= `20/65 + 36/65 `

= `56/65`

⇒ x + y = `sin ^-1(56/65)`

or, `cos^-1(12/13) + sin^-1 (3/5)`

= `sin^-1(56/65)`

Let `sin^(-1)  5/13` = x and `cos^(-1)  3/5` = y

⇒ sin x = `5/13 ` and cos y = `3/5`

or tan x = `5/12` and tan y = `4/3`

⇒ x = `tan^-1  5/12` and y = `tan^(-1)  4/3`

x + y = `tan^-1  5/12 + tan^-1  4/3`

= `tan^-1 ((5/12 + 4/3)/(1 - 5/12 xx 4/3))`

= `tan^(-1) ((15+48)/(36-20))`

= `tan^(-1)  63/16`

Let x = tan² θ

⇒ `sqrtx` = tan θ

⇒ θ = `tan^(-1) sqrtx`  ...(1)

∴ `(1-x)/(1+x)`

= `(1-tan^2 θ)/(1 + tan^2 θ)`

= cos 2θ

Now we have,

R.H.S = `1/2 cos^(-1)  (1-x)/(1+x)`

= `1/2 cos^(-1)(cos 2θ)`

= `1/2 xx 2θ`

= θ

= `tan^(-1) sqrtx`  ....[From (1)]

R.H.S. = L.H.S.

L.H.S. = `cot^(-1)  ((sqrt(1+sin x) + sqrt(1-sinx))/(sqrt(1+sin x) - sqrt(1- sinx)))`

= `cot^(-1)  (sqrt(1+sin x) + sqrt(1-sinx))/(sqrt(1+sin x) - sqrt(1 - sin x)) xx (sqrt(1+sin x) + sqrt(1-sinx))/(sqrt(1+sin x) - sqrt(1 - sin x))`

= `cot^(-1)  ((1+sinx) + (1-sinx) + 2sqrt(1 - sin^2 x))/((1+sinx) - (1 - sinx)`

= `cot^(-1)  (2(1 + cos x))/(2sin x)`

= `cot^(-1)  (1+ cosx)/sin x`

= `cot^(-1)  (2 cos^2  x/2)/(2sin  x/2 cos  x/2)`

= `cot^-1 (cot  x/2)`

= `x/2`

L.H.S. = R.H.S.

Let x = sin θ

Then, sin⁻¹ x = θ

We have

R.H.S = sin⁻¹ (3x − 4x³) = sin⁻¹ (3 sin θ − 4 sin³θ)

= sin⁻¹ (sin 3θ) = sin⁻¹ (3 sin θ − 4 sin³θ)

= 3θ = sin⁻¹ (3 sin θ − 4 sin³θ)

= 3 sin⁻¹ x = sin⁻¹ (3 sin θ − 4 sin³θ)

R.H.S = L.H.S

sin² A + cos² A = 1

1 + tan² A = sec² A 

1 + cot² A = cosec² A

For an acute angle A in a right-angled triangle:

• Hypotenuse is the side opposite the right angle.

• Perpendicular is the side opposite angle A.

• Base is the side adjacent to angle A.