0 mehakNewbie Asked: June 24, 20232023-06-24T20:29:19+05:30 2023-06-24T20:29:19+05:30In: CBSE 3. In fig. 5.37, find tan P and cot R. Is tan P = cot R? 0 Help with this question. Which formula used? Class 10th cbse rd sharma trigonometric identites rd sharma class 10thtrigonometric identities Share Facebook 1 Answer Voted Oldest Recent mehak Newbie 2023-06-25T19:48:20+05:30Added an answer on June 25, 2023 at 7:48 pm Solution: By using Pythagoras theorem in △PQR, we have PR^{2} = PQ^{2} + QR^{2} Putting the length of given side PR and PQ in the above equation, 13^{2 }= 12^{2} + QR^{2} QR^{2} = 13^{2} – 12^{2} QR^{2} = 169 – 144 QR^{2 }= 25 QR = √25 = 5 By definition, tan P = Perpendicular side opposite to P/ Base side adjacent to angle P tan P = QR/PQ tan P = 5/12 ………. (1) And, cot R= Base/Perpendicular cot R= QR/PQ cot R= 5/12 …. (2) When comparing equation (1) and (2), we can see that R.H.S of both the equation is equal. Therefore, L.H.S of both equations should also be equal. ∴ tan P = cot R Yes, tan P = cot R = 5/12 0 Reply Share Share Share on Facebook Share on Twitter Share on LinkedIn Share on WhatsApp Leave an answerLeave an answerCancel reply Featured image Select file Browse Add a Video to describe the problem better. Video type Youtube Vimeo Dailymotion Facebook Choose from here the video type. Video ID Put Video ID here: https://www.youtube.com/watch?v=sdUUx5FdySs Ex: "sdUUx5FdySs". Click on image to update the captcha. Save my name, email, and website in this browser for the next time I comment. Related Questions 16. A copper sphere of radius 3 cm is melted and recast into a right circular cone of ... 17. A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of ... 18. The diameters of the internal and external surfaces of a hollow spherical shell are 10cm and 6 ... 19. How many coins 1.75 cm in diameter and 2 mm thick must be melted to form a ... 20. The surface area of a solid metallic sphere is 616 cm2. It is melted and recast into ...

Solution:By using Pythagoras theorem in △PQR, we have

PR

^{2}= PQ^{2}+ QR^{2}Putting the length of given side PR and PQ in the above equation,

13

^{2 }= 12^{2}+ QR^{2}QR

^{2}= 13^{2}– 12^{2}QR

^{2}= 169 – 144QR

^{2 }= 25QR = √25 = 5

By definition,

tan P = Perpendicular side opposite to P/ Base side adjacent to angle P

tan P = QR/PQ

tan P = 5/12 ………. (1)

And,

cot R= Base/Perpendicular

cot R= QR/PQ

cot R= 5/12 …. (2)

When comparing equation (1) and (2), we can see that R.H.S of both the equation is equal.

Therefore, L.H.S of both equations should also be equal.

∴ tan P = cot R

Yes, tan P = cot R = 5/12