Trigonometric identity:
(1 + tan2θ)(1 - sin θ)(1 +sin θ) = 1
This trigonometric identity can, as the previous one, be solved in two ways:
- Convert all trigonometric functions to sine and cosine and simplify
- Apply identities other than that (more advanced method)
Recall the trigonometric identity 1 + tan2θ = sec2θ. Applying it to the LHS expression,
(1 + tan2θ)(1 - sin θ)(1 +sin θ)
= (sec2θ)(1 - sin θ)(1 +sin θ)The other part of the LHS, (1 - sin θ)(1 +sin θ), is in the form of (a + b)(a - b), which can be simplified to a2 - b2, so it can be simplified to 12 - sin2 θ,
= (sec2θ)(1 - sin2 θ)From the identity sin2 θ + cos2 θ = 1, it follows that cos2 θ = 1 - sin2 θ, applying this to the above expression,
= (sec2θ)(cos2 θ)We know that sec θ = 1/cos θ, hence sec2θ = 1/cos2θ. Hence the expression becomes,
= [ 1/cos2θ ] * cos2,
= 1... which is same as the RHS.
Trigonometric identities applied:
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