Solve-4-1-sin 2-1-tan 2-

The correct option is A 4 We know that (1 sin^2θ nbsp;=cos^2θ nbsp; and 1 + tan^2θ nbsp;= sec^2θ) 4[1 sin^2θ] [1 + tan^2θ ] = 4 cos^2θ× sec^2θ = 4 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Trigonometric Identities

Solve:4[1-sin...

Question

Solve:
$4 [1 - s i n^{2} θ] [1 + t a n^{2} θ$ ]

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
4

We know that
$(1 - s i n^{2} θ = c o s^{2} θ and 1 + t a n^{2} θ = s e c^{2} θ)$

$4 [1 - s i n^{2} θ] [1 + t a n^{2} θ] = 4 c o s^{2} θ \times s e c^{2} θ = 4$

Suggest Corrections

Similar questions

4 [1 – $s i n^{2} θ$ ] [1 + $t a n^{2} θ$ ] = ___.

Q. Prove

\frac{c o s^{2} θ + t a n^{2} θ - 1}{s i n^{2} θ} = t a n^{2} θ

[4 MARKS]

Q. Solve

t a n^{2} Θ - s i n^{2} Θ = t a n^{2} Θ s i n^{2} Θ

\frac{{cos}^{2} θ + {tan}^{2} θ - 1}{{sin}^{2} θ} = {tan}^{2} θ

\sqrt{1 + t a n^{2} θ} \sqrt{1 + c o t^{2} θ} \sqrt{1 - c o s^{2} θ} \sqrt{1 - s i n^{2} θ} =

Join BYJU'S Learning Program

Solve: 4[1–sin2θ][1+tan2θ]

The correct option is A 4 We know that (1−sin2θ=cos2θ and 1+tan2θ=sec2θ) 4[1−sin2θ][1+tan2θ]=4 cos2θ×sec2θ=4

Solve:
$4 [1 - s i n^{2} θ] [1 + t a n^{2} θ$ ]

The correct option is A
4

We know that
$(1 - s i n^{2} θ = c o s^{2} θ and 1 + t a n^{2} θ = s e c^{2} θ)$

$4 [1 - s i n^{2} θ] [1 + t a n^{2} θ] = 4 c o s^{2} θ \times s e c^{2} θ = 4$