Question

Solve (1+tan^2 theta)(1-sin theta)(1+sin theta) (1+cos theta)(1-cos theta) (1+cot^2 theta)

Answer 1

Here, We have given equation :

$\implies \bf (1 + Tan^{2} \: \theta ) (1 - sin \: \theta) (1 + sin \: \theta) (1 - cos \: \theta) (1 + cos \: \theta) (1 + cot^{2} \: \theta)$

Let's solve it :

Using Identities :

$\bullet \: \: {\boxed{\sf {Sin^{2} \: \theta + Cos^{2} \: \theta = 1}}}$

$\bullet \: \: {\boxed{\sf {1 - Sin^{2} \: \theta = Cos^{2} \: \theta }}}$

$\bullet \: \: {\boxed{\sf {Sin^{2} \: \theta = 1 - Cos^{2} \: \theta}}}$

Ratio Relation Used :

$\bullet \: \: {\boxed{\sf {Tan^{2} \: \theta = \dfrac{Sin^{2} \: \theta}{Cos^{2} \: \theta }}}}$

$\bullet \: \: {\boxed{\sf {Cot^{2} \: \theta = \dfrac{Cos^{2} \: \theta}{Sin^{2} \: \theta }}}}$

Now, solving it :

$\bullet \: \: {\sf {Tan^{2} \: \theta = \dfrac{Sin^{2} \: \theta}{Cos^{2} \: \theta }}}$

$\implies \sf (1 + Tan^{2} \: \theta ) (1 - sin \: \theta) (1 + sin \: \theta) (1 - cos \: \theta) (1 + cos \: \theta) (1 + cot^{2} \: \theta)$

$\implies \sf (1 + \dfrac{sin^{2} \: \theta}{cos^{2} \: \theta} ) (1 - sin \: \theta) (1 + sin \: \theta) (1 - cos \: \theta) (1 + cos \: \theta) (1 + cot^{2} \: \theta)$

$\implies \sf (1 + \dfrac{cos^{2} \: \theta + sin^{2} \: \theta}{cos^{2} \: \theta} ) (1 - sin \: \theta) (1 + sin \: \theta) (1 - cos \: \theta) (1 + cos \: \theta) (1 + cot^{2} \: \theta)$

Using Identity :

$\implies \sf (a + b)(a -b) = a^{2} - b^{2}$

$\implies \sf (1 + \dfrac{cos^{2} \: \theta + sin^{2} \: \theta}{cos^{2} \: \theta} ) (1)^{2} - (sin \: \theta)^{2} (1 - cos \: \theta) (1 + cos \: \theta) (1 + cot^{2} \: \theta)$

$\implies \sf (\dfrac{cos^{2} \: \theta + sin^{2} \: \theta}{cos^{2} \: \theta} ) (1 - sin^{2} \: \theta) (1 - cos \: \theta) (1 + cos \: \theta) (1 + cot^{2} \: \theta)$

$\therefore\sf (a + b)(a -b) = a^{2} - b^{2}$

$\implies \sf (\dfrac{cos^{2} \: \theta + sin^{2} \: \theta}{cos^{2} \: \theta} ) (1 - sin^{2} \: \theta) (1)^{2} - (cos \: \theta)^{2} (1 + cot^{2} \: \theta)$

$\implies \sf ( \dfrac{cos^{2} \: \theta + sin^{2} \: \theta}{cos^{2} \: \theta} ) (1 - sin^{2} \: \theta) 1 - (cos^{2} \: \theta) (1 + cot^{2} \: \theta)$

$\implies \sf (\dfrac{cos^{2} \: \theta + sin^{2} \: \theta}{cos^{2} \: \theta} ) (1 - sin^{2} \: \theta) 1 - (cos^{2} \: \theta) (1 + \dfrac{cos^{2} \: \theta}{sin^{2} \: \theta})$

$\bullet \: \: {\sf {Cot^{2} \: \theta = \dfrac{Cos^{2} \: \theta}{Sin^{2} \: \theta }}}$

$\implies \sf ( \dfrac{cos^{2} \: \theta + sin^{2} \: \theta}{cos^{2} \: \theta} ) (1 - sin^{2} \: \theta) 1 - (cos^{2} \: \theta) ( \dfrac{sin^{2} \: \theta + cos^{2} \: \theta}{sin^{2} \: \theta})$

$\implies \sf ( \dfrac{1}{cos^{2} \: \theta} ) (1 - sin^{2} \: \theta) 1 - (cos^{2} \: \theta) ( \dfrac{1}{sin^{2} \: \theta})$

$\bullet \: \: {\sf {1 - Sin^{2} \: \theta = Cos^{2} \: \theta }}$

$\implies \sf ( \dfrac{1}{cos^{2} \: \theta} ) (cos^{2} \: \theta) (1 - (cos^{2} \: \theta) ( \dfrac{1}{sin^{2} \: \theta})$

$\implies \sf (1) (1 - (cos^{2} \: \theta) ( \dfrac{1}{sin^{2} \: \theta})$

$\bullet \: \: {\sf {Sin^{2} \: \theta = 1 - Cos^{2} \: \theta}}$

$\implies \sf (1) (sin^{2} \: \theta) ( \dfrac{1}{sin^{2} \: \theta})$

$\implies \sf (1) \times (1)$

$\implies \sf 1$

${\boxed{\bf{Required \: Answer = {\red{1}}}}}$

Answer 2

Answer:

$(1 + tan ^{2} \theta)(1 - sin \theta)(1 + sin \theta)(1 + cos \theta)(1 - cos \theta)(1 + {cot}^{2} \theta) \\ \\ = (1 + {tan}^{2} \theta)(1 + {cot}^{2} \theta)(1 - {sin}^{2} \theta)(1 - {cos}^{2} \theta) \\ \\ = (1 + {tan}^{2 } \theta)(1 + {cot}^{2} \theta) {cos}^{2} \theta {sin}^{2} \theta \\ \\ = ( { \cos}^{2} \theta {sin}^{2} \theta+ { \cos}^{2} \theta {sin}^{2} \theta {cot}^{2} \theta + { \cos}^{2} \theta {sin}^{2} \theta {tan}^{2} \theta +{ \cos}^{2} \theta {sin}^{2} \theta ) \\ \\ = 2{ \cos}^{2} \theta {sin}^{2} \theta+ {cos}^{4} \theta + {sin}^{4} \\ \\ = {( {sin}^{2} \theta + {cos}^{2} \theta)}^{2} \\ \\ = {1}^{2} \\ \\ = 1$