Question

Tan (60°+theta).tan (60°-theta) = 2cos 2theta +1/2cos2theta -1

Answer 1

Step-by-step explanation:

Given  Tan (60°+theta).tan (60°-theta) = 2cos 2theta +1/2cos2theta -1

• So l. h.s will be tan(60 + theta) tan (60 – theta)
• So we can write this as tan (a + b) tan (a – b)
•      So (tan 60 + tan theta / 1 – tan 60 tan theta) (tan 60 – tan theta / 1 + tan 60 tan theta)
•      So (√3 + tan theta / 1 - √3 tan theta) (√3 – tan theta / 1 + √3 tan theta)
• So this will be (a + b)(a – b) = a^2 – b^2      
•      So (√3)^2 – (tan^2 theta) / 1 – (√3tan theta)^2
•             3 – tan^2 theta / 1 – 3 tan^2 theta
•              3 – sin^2 theta / cos^2 theta / 1 – 3 sin^2 theta / cos^2 theta
•               3 cos^2 theta – sin^2 theta / cos^2 theta – 3 sin^2 theta
•             So cos^2 theta – sin^^2 theta + 2 cos^2 theta / cos^2 theta – sin^2 theta – 2 sin^2 theta  
•                    Now cos 2theta = cos^2 theta – sin^2 theta
•                                               = 2 cos ^2 theta – 1
•                                                  = 1 – 2 sin^2 theta
•     So substituting we get
•                    So cos 2 theta + cos 2theta + 1 / cos 2 theta + cos 2 theta – 1
•                    So we get 2 cos 2 theta + 1 / 2 cos 2 theta - 1

Reference link will be

https://brainly.in/question/17956781

Answer 2

hope it helps you ....✨✨✨