Question

(tan²o + 1). Cas²o =1​

Answer 1

$\huge\boxed{\fcolorbox{cyan}{grey}{Solution:-}}$.

➡Proof Here ,

↪(Tan²0+1)(Cos²0)=1.

↪(Tan²0+1)(Cos²0)=1.➡Take L.H.S.

↪(Tan²0+1)(Cos²0)=1.➡Take L.H.S.= (Tan²0+1)(Cos²0).

[ We know , (Tan²x+1)=Sec²x ]

So,

= (Sec²0)(Cos²0).

= (Sec²0)(Cos²0).= (1/cos²0)(Cos²0).

= (Sec²0)(Cos²0).= (1/cos²0)(Cos²0).= 1.

That's Proved .

$\huge\boxed{\fcolorbox{cyan}{grey}{Hopes it's helps u:-}}$

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Answer 2

$\huge \red{ \mathfrak{solution}}$

To prove :

◆

$\boxed{( {tan}^{2} \alpha + 1) \times {cos}^{2} \alpha = 1}$

Take LHS :

$\leadsto \: ( {tan}^{2} \alpha + 1) \times {cos}^{2} \alpha$

As we know that :

$\blue{ \boxed{ \rightarrow \: {sec}^{2} \alpha - {tan}^{2} \alpha = 1}} \\ \\ \bold{ then} \\ \\ \green{\boxed{\rightarrow \: {sec}^{2} \alpha = 1 + {tan}^{2} \alpha }}$

now,

$\leadsto \: {sec}^{2} \alpha \times {cos}^{2} \alpha$

Again we know that :

$\pink{\boxed{cos \theta \: = \frac{1}{sec \theta} }}$

$\leadsto \: {sec}^{2} \alpha \times \frac{1}{ {sec}^{2} \alpha } \\ \\ \leadsto \: \cancel \frac{ {sec}^{2} \alpha }{ {sec}^{2} \alpha } \: 1 \\ \\ \leadsto \: 1$

LHS= RHS

HENCE ,

$\boxed{ \red{\boxed { \bold{proved}}}}$