L.H.S. =sinθ−cosθ+1sinθ+cosθ−1. =tanθ+secθ−1tanθ−secθ+1=(tanθ+secθ)−(sec2θ−tan2θ)tanθ−secθ+1. =(tanθ+secθ)−(secθ−tanθ)(secθ+tanθ)tanθ−secθ+1. =(tanθ+secθ)(1−sec ...

Given,. sinθ−2sin3θ2cos3θ−cosθ=tanθ. L.H.S=sinθ−2sin3θ2cos3θ−cosθ. =sinθ(1−2sin2θ)cosθ(2cos2θ−1). =sinθcosθ[1−2sin2θ2(1−sin2θ)−1]. =tanθ[1−2sin2θ2−2sin2θ−1]. = ...

25 окт. 2024 г. · Given: [(sin θ tan θ + cos θ )2 - 1 ] Formula used: tan θ = sin θ / cos θ, Sec²θ - 1 = Tan²θ,

Prove: sin θ(1 + tan θ) + cos θ(1 + cot θ) = (sec θ + cosec θ) . · Prove the identity (sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θ · If ((sinθ−cosecθ)(cosθ− ...

Prove the following trigonometric identities. (i) 1+cosθ+sinθ1+cosθ−sinθ=1+sinθcosθ. (ii) sinθ−cosθ+1sinθ+cosθ−1=1secθ−tanθ. (iii) cosθ−sinθ+1cosθ+sinθ−1=cosecθ ...

Because the two sides have been shown to be equivalent, the equation is an identity. sin(θ)tan(θ)+cos(θ)=sec(θ) is an identity.

The domain is all the values of θ that can be plugged into the function. sin(θ), θ can be any angle cos(θ), θ can be any angle tan(θ) ...

Detailed step by step solution for prove (sin(2θ)+sin(θ))/(cos(2θ)+cos(θ)+1)=tan(θ)

Solution · LHS=(1−sinθ)(1+sinθ)=1−sin2θ(difference of two squares)=cos2θ(Pythagorean identity)=RHS · LHS=2cos3θ−cosθsinθcos2θ−sin3θ=cosθ(2cos2θ−1)sinθ(cos2θ−sin2θ) ...

Let us consider a right Δ ABC right angled at B. Now it is given that tanθ=ABBC=2021. So, if AB = 20k, then BC = 21k where k is a positive number.