Question

secθ(1+sinθ/cosθ+cosθ/1+sinθ)-2tan²θ​

Answer 1

Solution−

Given expression is

$\rm :\longmapsto\:sec\theta \bigg( \dfrac{1 + sin\theta }{cos\theta } + \dfrac{cos\theta }{1 + sin\theta } \bigg) - 2 \: {tan}^{2} \theta$

$\rm \: = \: \:sec\theta \bigg(\dfrac{ {(1 + sin\theta )}^{2} + {cos}^{2}\theta }{cos\theta (1 + sin\theta )} \bigg) - 2 {tan}^{2}\theta$

We know,

$\boxed{ \bf{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

So using this identity, we get

$\rm \: = \: \:sec\theta \bigg(\dfrac{1 + {sin}^{2} \theta + 2sin\theta + {cos}^{2}\theta }{cos\theta (1 + sin\theta )} \bigg) - 2 {tan}^{2}\theta$

$\rm \: = \: \:sec\theta \bigg(\dfrac{1 + {sin}^{2} \theta + {cos}^{2}\theta + 2sin\theta }{cos\theta (1 + sin\theta )} \bigg) - 2 {tan}^{2}\theta$

We know that,

$\boxed{ \bf{ \: {sin}^{2}x + {cos}^{2}x = 1}}$

So, using this, we have

$\rm \: = \: \:sec\theta \bigg(\dfrac{1 + 1 + 2sin\theta }{cos\theta (1 + sin\theta )} \bigg) - 2 {tan}^{2}\theta$

$\rm \: = \: \:sec\theta \bigg(\dfrac{2 + 2sin\theta }{cos\theta (1 + sin\theta )} \bigg) - 2 {tan}^{2}\theta$

$\rm \: = \: \:sec\theta \bigg(\dfrac{2 }{cos\theta } \bigg) - 2 {tan}^{2}\theta$

$\rm \: = \: \:sec\theta (2sec\theta ) - 2 {tan}^{2} \theta$

$\rm \: = \: \: {2sec}^{2}\theta - {2tan}^{2} \theta$

$\rm \: = \: \:2( {sec}^{2}\theta - {tan}^{2} \theta )$

We know,

$\boxed{ \bf{ \: {sec}^{2} x - {tan}^{2} x = 1}}$

• So, using this, we have

$\rm \: = \: \:2 \times 1$

$\rm \: = \: \:2$

Hence,

The value of

$\boxed{ \bf{ \: \:sec\theta \bigg( \dfrac{1 + sin\theta }{cos\theta } + \dfrac{cos\theta }{1 + sin\theta } \bigg) - 2 \: {tan}^{2} \theta \: = \: 2}}$

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Answer 2

Answer:

Solution−

Given expression is

\rm :\longmapsto\:sec\theta \bigg( \dfrac{1 + sin\theta }{cos\theta } + \dfrac{cos\theta }{1 + sin\theta } \bigg) - 2 \: {tan}^{2} \theta:⟼secθ(

cosθ

1+sinθ

+

1+sinθ

cosθ

)−2tan

2

θ

\rm \: = \: \:sec\theta \bigg(\dfrac{ {(1 + sin\theta )}^{2} + {cos}^{2}\theta }{cos\theta (1 + sin\theta )} \bigg) - 2 {tan}^{2}\theta=secθ(

cosθ(1+sinθ)

(1+sinθ)

2

+cos

2

θ

)−2tan

2

θ

We know,

\boxed{ \bf{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}

(x+y)

2

=x

2

+y

2

+2xy

So using this identity, we get

\rm \: = \: \:sec\theta \bigg(\dfrac{1 + {sin}^{2} \theta + 2sin\theta + {cos}^{2}\theta }{cos\theta (1 + sin\theta )} \bigg) - 2 {tan}^{2}\theta=secθ(

cosθ(1+sinθ)

1+sin

2

θ+2sinθ+cos

2

θ

)−2tan

2

θ

\rm \: = \: \:sec\theta \bigg(\dfrac{1 + {sin}^{2} \theta + {cos}^{2}\theta + 2sin\theta }{cos\theta (1 + sin\theta )} \bigg) - 2 {tan}^{2}\theta=secθ(

cosθ(1+sinθ)

1+sin

2

θ+cos

2

θ+2sinθ

)−2tan

2

θ

We know that,

\boxed{ \bf{ \: {sin}^{2}x + {cos}^{2}x = 1}}

sin

2

x+cos

2

x=1

So, using this, we have

\rm \: = \: \:sec\theta \bigg(\dfrac{1 + 1 + 2sin\theta }{cos\theta (1 + sin\theta )} \bigg) - 2 {tan}^{2}\theta=secθ(

cosθ(1+sinθ)

1+1+2sinθ

)−2tan

2

θ

\rm \: = \: \:sec\theta \bigg(\dfrac{2 + 2sin\theta }{cos\theta (1 + sin\theta )} \bigg) - 2 {tan}^{2}\theta=secθ(

cosθ(1+sinθ)

2+2sinθ

)−2tan

2

θ

\rm \: = \: \:sec\theta \bigg(\dfrac{2 }{cos\theta } \bigg) - 2 {tan}^{2}\theta=secθ(

cosθ

2

)−2tan

2

θ

\rm \: = \: \:sec\theta (2sec\theta ) - 2 {tan}^{2} \theta=secθ(2secθ)−2tan

2

θ

\rm \: = \: \: {2sec}^{2}\theta - {2tan}^{2} \theta=2sec

2

θ−2tan

2

θ

\rm \: = \: \:2( {sec}^{2}\theta - {tan}^{2} \theta )=2(sec

2

θ−tan

2

θ)

We know,

\boxed{ \bf{ \: {sec}^{2} x - {tan}^{2} x = 1}}

sec

2

x−tan

2

x=1

So, using this, we have

\rm \: = \: \:2 \times 1=2×1

\rm \: = \: \:2=2

Hence,

The value of

\boxed{ \bf{ \: \:sec\theta \bigg( \dfrac{1 + sin\theta }{cos\theta } + \dfrac{cos\theta }{1 + sin\theta } \bigg) - 2 \: {tan}^{2} \theta \: = \: 2}}

secθ(

cosθ

1+sinθ

+

1+sinθ

cosθ

)−2tan

2

θ=2

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