Evaluate cos 48° cos 42° − sin 48° sin 42°
Note : This is similar to Ex 8.3, 2 (ii) of NCERT – Chapter 8 Class 10
Check the answer here https://www.teachoo.com/1774/531/Ex-8.3--1---Evaluate-(i)-sin-18---cos-72-(ii)-tan-26-cot-64/category/Ex-8.3/
CBSE Class 10 Sample Paper for 2020 Boards - Maths Basic
CBSE Class 10 Sample Paper for 2020 Boards - Maths Basic
Last updated at December 16, 2024 by Teachoo
Note : This is similar to Ex 8.3, 2 (ii) of NCERT – Chapter 8 Class 10
Check the answer here https://www.teachoo.com/1774/531/Ex-8.3--1---Evaluate-(i)-sin-18---cos-72-(ii)-tan-26-cot-64/category/Ex-8.3/
Transcript
Question 24 (OR 2nd question) Evaluate cos 48° cos 42° − sin 48° sin 42° cos 48° cos 42° – sin 48° sin 42° = cos (90 – 42)° cos (90 – 48)° – sin 48° sin 42° = sin 42° sin 48° – sin 48° sin 42° = 0 (cos (90 – θ) = sin θ )