0 mehakNewbie Asked: June 24, 20232023-06-24T20:39:22+05:30 2023-06-24T20:39:22+05:30In: CBSE 7. If cot θ = 7/8, evaluate (i) (1+sin θ)(1–sin θ)/ (1+cos θ)(1–cos θ) (ii) cot2 θ 0 Explain the question. Justify the answer Class 10th, rd sharma, trigonometric identities rd sharma class 10thtrigonometric identities Share Facebook 1 Answer Voted Oldest Recent mehak Newbie 2023-06-25T19:50:42+05:30Added an answer on June 25, 2023 at 7:50 pm Solution: (i) Required to evaluate: , given = cot θ = 7/8 Taking the numerator, we have (1+sin θ)(1–sin θ) = 1 – sin^{2} θ [Since, (a+b)(a-b) = a^{2} – b^{2}] Similarly, (1+cos θ)(1–cos θ) = 1 – cos^{2} θ We know that, sin^{2} θ + cos^{2} θ = 1 ⇒ 1 – cos^{2} θ = sin^{2} θ And, 1 – sin^{2} θ = cos^{2} θ Thus, (1+sin θ)(1 –sin θ) = 1 – sin^{2} θ = cos^{2} θ (1+cos θ)(1–cos θ) = 1 – cos^{2} θ = sin^{2} θ ⇒ = cos^{2} θ/ sin^{2} θ = (cos θ/sin θ)^{2} And, we know that (cos θ/sin θ) = cot θ ⇒ = (cot θ)^{2} = (7/8)^{2} = 49/ 64 (ii) Given, cot θ = 7/8 So, by squaring on both sides we get (cot θ)^{2} = (7/8)^{2} ∴ cot θ^{2} = 49/64 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:(i)Required to evaluate:, given = cot θ = 7/8

Taking the numerator, we have

(1+sin θ)(1–sin θ) = 1 – sin

^{2}θ [Since, (a+b)(a-b) = a^{2}– b^{2}]Similarly,

(1+cos θ)(1–cos θ) = 1 – cos

^{2}θWe know that,

sin

^{2}θ + cos^{2}θ = 1⇒ 1 – cos

^{2}θ = sin^{2}θAnd,

1 – sin

^{2}θ = cos^{2}θThus,

(1+sin θ)(1 –sin θ) = 1 – sin

^{2}θ = cos^{2}θ(1+cos θ)(1–cos θ) = 1 – cos

^{2}θ = sin^{2}θ⇒

= cos

^{2}θ/ sin^{2}θ= (cos θ/sin θ)

^{2}And, we know that (cos θ/sin θ) = cot θ

⇒

=(cot θ)^{2}= (7/8)

^{2}= 49/ 64

(ii)Given,cot θ = 7/8

So, by squaring on both sides we get

(cot θ)

^{2}= (7/8)^{2}∴ cot θ

^{2}= 49/64