0 [email protected]Scholar Asked: April 8, 20212021-04-08T00:28:57+05:30 2021-04-08T00:28:57+05:30In: CBSE Without using trigonometric tables, prove that: sin35°sin55° – cos35°cos55° = 0 0 This question is from trigonometry topic – trigonometric ratios on complementary angles We have been asked to prove that sin35°sin55° – cos35°cos55° = 0 without using trigonometric tables RS Aggarwal, class 10, chapter 12, question no 3(iv) class 10 cbsers aggarwaltrigonometric ratios Share Facebook Related Questions Prove that: (secθ+tanθ)/(secθ-tanθ) = (secθ+tanθ)² = 1+2tan²θ+2secθtanθ Prove that: (cosecθ + cotθ)/(cosecθ - cotθ) = (cosecθ + cotθ)² = 1 + 2cot²θ + 2cosecθcotθ Prove that: (1+cosθ-sin²θ)/sinθ(1+cosθ) = cotθ Prove that: (sinθ+cosθ)/(sinθ-cosθ) + (sinθ-cosθ)/(sinθ+cosθ) = 2/(1-2cos²θ) Prove that: (sinθ-cosθ)/(sinθ+cosθ) + (sinθ+cosθ)/(sinθ-cosθ) = 2/(sin²θ-1) 1 Answer Voted Oldest Recent MathsMentor Guru 2021-04-10T17:28:41+05:30Added an answer on April 10, 2021 at 5:28 pm We know that, sin (90 – θ) = cos θ So, the given can be expressed as sin (90 – 55)° sin (90 – 35)° – cos 35° cos 55° = cos 55° cos 35° – cos 35° cos 55° = 0 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.

We know that, sin (90 – θ) = cos θ So, the given can be expressed as sin (90 – 55)° sin (90 – 35)° – cos 35° cos 55°

= cos 55° cos 35° – cos 35° cos 55°

= 0