By Applying Pythagoras theorem in ΔPQR , we get

PR^{2} = PQ^{2} + QR^{2} = (13)^{2} = (12)^{2} + QR^{2 }=^{ }169 = 144 + QR^{2}

⇒ QR^{2} = 25 ⇒ QR = 5 cm

Now,

tan P = QR/PQ = 5/12

cot R = QR/PQ = 5/12

A/q

tan P – cot R = 5/12 - 5/12 = 0

Let ΔABC be a right-angled triangle, right-angled at B.

We know that sin A = BC/AC = 3/4

Let BC be 3k and AC will be 4k where k is a positive real number.

By Pythagoras theorem we get,

AC^{2} = AB^{2} + BC^{2 }

(4k)^{2} = AB^{2} + (3k)^{2}

16k^{2} - 9k^{2} = AB^{2}

AB^{2 }=^{ }7k^{2}

AB = √7 k

cos A = AB/AC = √7 k/4k = √7/4

tan A = BC/AB = 3k/√7 k = 3/√7

Let ΔABC be a right-angled triangle, right-angled at B.

We know that cot A = AB/BC = 8/15 (Given)

Let AB be 8k and BC will be 15k where k is a positive real number.

By Pythagoras theorem we get,

AC^{2} = AB^{2} + BC^{2 }

AC^{2} = (8k)^{2} + (15k)^{2}

AC^{2} = 64k^{2} + 225k^{2}

AC^{2}^{ }=^{ }289k^{2}

AC = 17 k

sin A = BC/AC = 15k/17k = 15/17

sec A = AC/AB = 17k/8 k = 17/8

Let ΔABC be a right-angled triangle, right-angled at B.

We know that sec θ = OP/OM = 13/12 (Given)

Let OP be 13k and OM will be 12k where k is a positive real number.

By Pythagoras theorem we get,

OP^{2} = OM^{2} + MP^{2 }

(13k)^{2} = (12k)^{2 }+ MP^{2 }

169k^{2} - 144k^{2} = MP^{2}

MP^{2}^{ }=^{ }25k^{2}

MP = 5

Now,

sin θ = MP/OP = 5k/13k = 5/13

cos θ = OM/OP = 12k/13k = 12/13

tan θ = MP/OM = 5k/12k = 5/12

cot θ = OM/MP = 12k/5k = 12/5

cosec θ = OP/MP = 13k/5k = 13/5