Simplification of 1-tan square a of COS square a

Simplification of 1-tan square a of COS square a

The original formula = 1 / cos? A-SiN? 2 / cos? A
=(1-sin²a)/cos²a
=cos²a/cos²a
=1

Help me simplify, cos square a times Tan square a + cos square a

cos^2atan^2a+cos^2a
=sin^2a+cos^2a
=1

Simplify cos squarex (1-tan square x)

Cos square x (1-tan square x)
=cos²x· (1-sin²x/cos²x)
=cos²x-sin²x
=cos2x

It is known that Tan θ = 1 3, then Cos2 θ + 1 2sin2θ= ___ .

∵tanθ=1
3，
∴cos2θ+1
2sin2θ=cos2θ+sinθcosθ
cos2θ+sin2θ=1+tanθ
1+tan2θ=1+1
Three
1+1
9=6
5，
So the answer is: 6
5．

Given Tan α = 1, find the value of 2sin2 α - 3sin α cos-5cos2 α Step by step oh, please, lo~

∵tana=1∴ cosa=sina∴ 2sin2α-3sinαcosα-5cos2α=(2sin²a-3sinacosa-5cos²a)/(sin²a+cos²a)=(2sin²a-3sin²a-5sin²a)/(sin²a+sin²a)= -6sin²a/(2sin²a)...

If 1 / Tan α = 3, cos 2 α + 1 / 2sian2 α =? Ask for detailed explanation

cos²α+1/2sian2α
=(cos²α+sinacosa)/(cos^2a+sin^2a)
=(1+cota)/(1+cot^2a)
=2/5

Simplification: 1-tan α - 1 + Tan α 1

(1 / - 1 + Tana) / 1-tana (the numerator is represented in brackets)
=[1/-（1-tana)]/1-tana
=-1 / (1-tana) 2 (multiply by 1-tana)
=-1 / 1-2tana + Tana (complete square formula)
If the original question is correct, it should be

Simplification: 1 + Tan α / 1-tan α

1+tanα/1-tanα
=[tan(π/4)+tanα]/1-tanαtan(π/4)
=tan(α+π/4)

Simplify Tan + 1 / Tan

tana＋1/tana
=sina/cosa+coa/sina
=(sina²+cosa²)/sinacosa
=1/sinacosa
=2/sin2a

Simplify Tan ((π / 12) + 1) / ((π / 12) - 1) for guidance,

The title is wrong
It could be
[ tan(π/12) +1 ] / [ tan(π/12) - 1 ]
= [ tan(π/12) + tan(π/4) ] / [ tan(π/12)*tan(π/4) - 1 ]
= - tan(π/12 + π/4)
= - tan(π/3)
= - √3